SMU PHYSICS 1303: Introduction to Mechanics

Stephen Sekula1

1Southern Methodist University Dallas, TX, USA

SPRING, 2019

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 1 Outline

Rotational Energy and Force Concepts

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 2 Rotational Energy and Force Concepts

Rotational Energy and Force Concepts

“Jet Engine” by William Warby, available via Creative Commons from Flickr S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 3 Rotational Energy and Force Concepts Key Ideas

The key ideas that we will explore in this section of the course are as follows: I We will begin to think about energy and force concepts as they relate to extended, rigid, rotating bodies. I We will see that there are rotational analogs to linear concepts,

like kinetic energy. “Sprunger Table Saw blade” by I We will come to understand how forces result in rotations, and lungstruck, available under Creative explore the concept of a torque. Commons from Flickr I We will connect the force and energy concepts through work, and complete a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 4 Rotational Energy and Force Concepts Key Ideas

The key ideas that we will explore in this section of the course are as follows: I We will begin to think about energy and force concepts as they relate to extended, rigid, rotating bodies. I We will see that there are rotational analogs to linear concepts,

like kinetic energy. “Sprunger Table Saw blade” by I We will come to understand how forces result in rotations, and lungstruck, available under Creative explore the concept of a torque. Commons from Flickr I We will connect the force and energy concepts through work, and complete a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 4 Rotational Energy and Force Concepts Key Ideas

The key ideas that we will explore in this section of the course are as follows: I We will begin to think about energy and force concepts as they relate to extended, rigid, rotating bodies. I We will see that there are rotational analogs to linear concepts,

like kinetic energy. “Sprunger Table Saw blade” by I We will come to understand how forces result in rotations, and lungstruck, available under Creative explore the concept of a torque. Commons from Flickr I We will connect the force and energy concepts through work, and complete a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 4 Rotational Energy and Force Concepts Key Ideas

The key ideas that we will explore in this section of the course are as follows: I We will begin to think about energy and force concepts as they relate to extended, rigid, rotating bodies. I We will see that there are rotational analogs to linear concepts,

like kinetic energy. “Sprunger Table Saw blade” by I We will come to understand how forces result in rotations, and lungstruck, available under Creative explore the concept of a torque. Commons from Flickr I We will connect the force and energy concepts through work, and complete a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 4 Rotational Energy and Force Concepts Key Ideas

The key ideas that we will explore in this section of the course are as follows: I We will begin to think about energy and force concepts as they relate to extended, rigid, rotating bodies. I We will see that there are rotational analogs to linear concepts,

like kinetic energy. “Sprunger Table Saw blade” by I We will come to understand how forces result in rotations, and lungstruck, available under Creative explore the concept of a torque. Commons from Flickr I We will connect the force and energy concepts through work, and complete a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 4 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Let me begin with a question: I am going to throw the vent fan pictured left at your hand at a linear speed of 170km/h. Is that okay? To which you probably replied: ‘No. That is highly unsafe.” Why? Because this giant vent fan would have a huge linear kinetic energy, part of which would be transferred to you either in an elastic or inelastic collision. That will hurt. A LOT. “Giant Ventilation Fan” by Christoper and available under Creative Commons from Flickr

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 5 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Let me begin with a question: I am going to throw the vent fan pictured left at your hand at a linear speed of 170km/h. Is that okay? To which you probably replied: ‘No. That is highly unsafe.” Why? Because this giant vent fan would have a huge linear kinetic energy, part of which would be transferred to you either in an elastic or inelastic collision. That will hurt. A LOT. “Giant Ventilation Fan” by Christoper and available under Creative Commons from Flickr

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 5 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Let me begin with a question: I am going to throw the vent fan pictured left at your hand at a linear speed of 170km/h. Is that okay? To which you probably replied: ‘No. That is highly unsafe.” Why? Because this giant vent fan would have a huge linear kinetic energy, part of which would be transferred to you either in an elastic or inelastic collision. That will hurt. A LOT. “Giant Ventilation Fan” by Christoper and available under Creative Commons from Flickr

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 5 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Let me begin with a question: I am going to throw the vent fan pictured left at your hand at a linear speed of 170km/h. Is that okay? To which you probably replied: ‘No. That is highly unsafe.” Why? Because this giant vent fan would have a huge linear kinetic energy, part of which would be transferred to you either in an elastic or inelastic collision. That will hurt. A LOT. “Giant Ventilation Fan” by Christoper and available under Creative Commons from Flickr

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 5 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Fine. Let me ask a different question: The vent fan has a uniform distribution of mass, so its center of mass (COM) is located dead center — just on its axis of rotation. I’m going to spin the fan so the tip of its blades move at 170km/h. But since the COM is still, it has no linear kinetic energy. I’m going to stick your hand into the path of the blades. Is that okay? To which you probably replied: ‘HECK no. That is highly unsafe.” Why? Because even though it’s “Giant Ventilation Fan” by Christoper and true that the COM has no linear kinetic energy, available under Creative Commons from Flickr those blades are moving, and there is clearly some kind of kinetic energy associated with rotation. Let’s figure out what that is.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 6 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Fine. Let me ask a different question: The vent fan has a uniform distribution of mass, so its center of mass (COM) is located dead center — just on its axis of rotation. I’m going to spin the fan so the tip of its blades move at 170km/h. But since the COM is still, it has no linear kinetic energy. I’m going to stick your hand into the path of the blades. Is that okay? To which you probably replied: ‘HECK no. That is highly unsafe.” Why? Because even though it’s “Giant Ventilation Fan” by Christoper and true that the COM has no linear kinetic energy, available under Creative Commons from Flickr those blades are moving, and there is clearly some kind of kinetic energy associated with rotation. Let’s figure out what that is.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 6 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Fine. Let me ask a different question: The vent fan has a uniform distribution of mass, so its center of mass (COM) is located dead center — just on its axis of rotation. I’m going to spin the fan so the tip of its blades move at 170km/h. But since the COM is still, it has no linear kinetic energy. I’m going to stick your hand into the path of the blades. Is that okay? To which you probably replied: ‘HECK no. That is highly unsafe.” Why? Because even though it’s “Giant Ventilation Fan” by Christoper and true that the COM has no linear kinetic energy, available under Creative Commons from Flickr those blades are moving, and there is clearly some kind of kinetic energy associated with rotation. Let’s figure out what that is.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 6 Rotational Energy and Force Concepts The Kinetic Energy of a Rotating Body

Fine. Let me ask a different question: The vent fan has a uniform distribution of mass, so its center of mass (COM) is located dead center — just on its axis of rotation. I’m going to spin the fan so the tip of its blades move at 170km/h. But since the COM is still, it has no linear kinetic energy. I’m going to stick your hand into the path of the blades. Is that okay? To which you probably replied: ‘HECK no. That is highly unsafe.” Why? Because even though it’s “Giant Ventilation Fan” by Christoper and true that the COM has no linear kinetic energy, available under Creative Commons from Flickr those blades are moving, and there is clearly some kind of kinetic energy associated with rotation. Let’s figure out what that is.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 6 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

Consider again the bike wheel we looked at in the lecture on describing rotational motion. Imagine it is spinning. Imagine also it is made of a huge number of bits of mass, each at some radius ri from the center of rotation and each with mass mi , where i labels all N (e.g. Avagadro’s Number) bits of mass. Choose one point (blue), and call it i = 1. It has linear kinetic 1 2 energy K1 = 2 m1v1 , where v1 is its linear speed. Choose a second point, i = 2 (red) and write its 1 2 kinetic energy: K2 = 2 m2v2 . Choose all the other points that make up the bike wheel, and keep calculating. . .

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 7 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

In fact, we can write this activity as one large sum of all the individual kinetic energies, a large sum that represents the total kinetic energy of the spinning wheel, which is just a big collection of points each with mass, mi , and linear speed, vi :

N N X X 1 K = K + K + ... + K = K = m v 2 total 1 2 N i 2 i i i=1 i=1 This might at first seem hopeless, but remember an important fact about points on a rigid rotating body: they all have the exact same angular speed, ω. In other words, v1/r1 = v2/r2 = ... = vN /rN = ω. Let’s use this!

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 8 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

N X 1 K = m v 2 total 2 i i i=1 N X 1 = m (ωr )2 2 i i i=1 N ! 1 X = m r 2 ω2 2 i i i=1

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 9 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

N X 1 K = m v 2 total 2 i i i=1 N X 1 = m (ωr )2 2 i i i=1 N ! 1 X = m r 2 ω2 2 i i i=1

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 9 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

N X 1 K = m v 2 total 2 i i i=1 N X 1 = m (ωr )2 2 i i i=1 N ! 1 X = m r 2 ω2 2 i i i=1

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 9 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

N ! 1 X K = m r 2 ω2 total 2 i i i=1

The complex rotational body made of a huge number of points has its kinetic energy more easily summarized by this equation:

1 K = Iω2 rotation 2

PN 2 where I = i=1 mi ri is known as the moment of inertia and summarizes how all the masses of each piece of the total contributes overall to the kinetic energy of rotation through their inertia.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 10 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

N ! 1 X K = m r 2 ω2 total 2 i i i=1

The complex rotational body made of a huge number of points has its kinetic energy more easily summarized by this equation:

1 K = Iω2 rotation 2

PN 2 where I = i=1 mi ri is known as the moment of inertia and summarizes how all the masses of each piece of the total contributes overall to the kinetic energy of rotation through their inertia.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 10 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

N ! 1 X K = m r 2 ω2 total 2 i i i=1

The complex rotational body made of a huge number of points has its kinetic energy more easily summarized by this equation:

1 K = Iω2 rotation 2

PN 2 where I = i=1 mi ri is known as the moment of inertia and summarizes how all the masses of each piece of the total contributes overall to the kinetic energy of rotation through their inertia.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 10 Rotational Energy and Force Concepts Linear Kinetic Energy of Every Point in a Rotating Body

N ! 1 X K = m r 2 ω2 total 2 i i i=1

The complex rotational body made of a huge number of points has its kinetic energy more easily summarized by this equation:

1 K = Iω2 rotation 2

PN 2 where I = i=1 mi ri is known as the moment of inertia and summarizes how all the masses of each piece of the total contributes overall to the kinetic energy of rotation through their inertia.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 10 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Point of Mass m circling a distance r from the center This is the easiest to calculate. There is one mass, m, all concentrated at a point a distance r from the center of rotation. Thus:

N X 2 2 Ipoint mass = mi ri = mr i=1 There is only one term in the sum.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 11 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Point of Mass m circling a distance r from the center This is the easiest to calculate. There is one mass, m, all concentrated at a point a distance r from the center of rotation. Thus:

N X 2 2 Ipoint mass = mi ri = mr i=1 There is only one term in the sum.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 11 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Point of Mass m circling a distance r from the center This is the easiest to calculate. There is one mass, m, all concentrated at a point a distance r from the center of rotation. Thus:

N X 2 2 Ipoint mass = mi ri = mr i=1 There is only one term in the sum.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 11 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Hoop of Mass m rotating a distance r about the z-axis This one requires some calculus. The hoop can be treated as made from an infinite number of infinitesimal points of mass, each with mass dm. All points of mass dm, however, are located at precisely the same distance from the z-axis. Thus:

N Z m m X 2 2 2 2 Ihoop about z = mi ri −→ r dm = mr = mr 0 i=1 0

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 12 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Hoop of Mass m rotating a distance r about the z-axis This one requires some calculus. The hoop can be treated as made from an infinite number of infinitesimal points of mass, each with mass dm. All points of mass dm, however, are located at precisely the same distance from the z-axis. Thus:

N Z m m X 2 2 2 2 Ihoop about z = mi ri −→ r dm = mr = mr 0 i=1 0

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 12 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Hoop of Mass m rotating a distance r about the z-axis This one requires some calculus. The hoop can be treated as made from an infinite number of infinitesimal points of mass, each with mass dm. All points of mass dm, however, are located at precisely the same distance from the z-axis. Thus:

N Z m m X 2 2 2 2 Ihoop about z = mi ri −→ r dm = mr = mr 0 i=1 0

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 12 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Hoop of Mass m rotating a distance r about the z-axis This one requires some calculus. The hoop can be treated as made from an infinite number of infinitesimal points of mass, each with mass dm. All points of mass dm, however, are located at precisely the same distance from the z-axis. Thus:

N Z m m X 2 2 2 2 Ihoop about z = mi ri −→ r dm = mr = mr 0 i=1 0

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 12 Rotational Energy and Force Concepts Moments of Inertia

The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Rod of Mass m and length L rotating about its center This one also requires some calculus, but I will skip it and quote the result:

1 I = mL2 rod about center 12

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 13 Rotational Energy and Force Concepts Moments of Inertia The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. A Rod of Mass m and length L rotating about its end Compare the previous case (a rod rotating about its center, like twirling a baton about its middle) to the case of taking the same rod and rotating it about its end (like using the baton as a bat instead): 1 I = mL2 rod about center 3 Note that for the same object, this has more moment of inertia, and thus for the same angular speed it would have more rotational kinetic energy. This shows that rotating the same object differently, even at the same angular speed, can lead to very different outcomes in energy.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 14 Rotational Energy and Force Concepts Moments of Inertia The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. Generally speaking: more mass located farther from the center of rotations means more moment of inertia, and more kinetic energy for the same angular speed. But, also generally speaking, that means that for two objects of the same mass, but one with a larger moment of inertia, more energy is required to accelerate the body with the larger moment of inertia to the same angular speed.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 15 Rotational Energy and Force Concepts Moments of Inertia The kinetic energy of a rotating, extended, and rigid body is easily determined once one has determined the moment of inertia of that specific body. Here are some examples of moments of inertia for common, simple objects. Generally speaking: more mass located farther from the center of rotations means more moment of inertia, and more kinetic energy for the same angular speed. But, also generally speaking, that means that for two objects of the same mass, but one with a larger moment of inertia, more energy is required to accelerate the body with the larger moment of inertia to the same angular speed.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 15 Rotational Energy and Force Concepts Using Forces to Rotate Bodies about an Axis We’ve been talking about spinning up objects so that they rotate about some axis (some fixed line in space), but we’ve avoided talking about how we get them spinning at some angular speed in the first place. I We know from Newton’s Second Law of Motion that if we want to take an object from a state of motion (e.g. zero angular speed) to a different state of motion (e.g. non-zero angular speed), we must apply a force. I Forces, applied to bodies that can rotate about an axis, that cause a rotation to occur result in what are known as torques. They are the rotational analog of linear force. I We know a few things about torques from our experiences with opening doors: I If the force is applied parallel to radial line that points back to the axis of rotation (either toward the hinges of the door or away from the hinges of the door), then no rotation of the door can result. I If, instead, there is a component of the force that is perpendicular to that radial line, a rotation is possible. I If we apply the same force closer to the hinges, it results in less rotation per unit time than if we apply the same force farther from the hinges (this is why door handles are most effective when placed as far as possible from the hinges)

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 16 Rotational Energy and Force Concepts Using Forces to Rotate Bodies about an Axis We’ve been talking about spinning up objects so that they rotate about some axis (some fixed line in space), but we’ve avoided talking about how we get them spinning at some angular speed in the first place. I We know from Newton’s Second Law of Motion that if we want to take an object from a state of motion (e.g. zero angular speed) to a different state of motion (e.g. non-zero angular speed), we must apply a force. I Forces, applied to bodies that can rotate about an axis, that cause a rotation to occur result in what are known as torques. They are the rotational analog of linear force. I We know a few things about torques from our experiences with opening doors: I If the force is applied parallel to radial line that points back to the axis of rotation (either toward the hinges of the door or away from the hinges of the door), then no rotation of the door can result. I If, instead, there is a component of the force that is perpendicular to that radial line, a rotation is possible. I If we apply the same force closer to the hinges, it results in less rotation per unit time than if we apply the same force farther from the hinges (this is why door handles are most effective when placed as far as possible from the hinges)

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 16 Rotational Energy and Force Concepts Using Forces to Rotate Bodies about an Axis We’ve been talking about spinning up objects so that they rotate about some axis (some fixed line in space), but we’ve avoided talking about how we get them spinning at some angular speed in the first place. I We know from Newton’s Second Law of Motion that if we want to take an object from a state of motion (e.g. zero angular speed) to a different state of motion (e.g. non-zero angular speed), we must apply a force. I Forces, applied to bodies that can rotate about an axis, that cause a rotation to occur result in what are known as torques. They are the rotational analog of linear force. I We know a few things about torques from our experiences with opening doors: I If the force is applied parallel to radial line that points back to the axis of rotation (either toward the hinges of the door or away from the hinges of the door), then no rotation of the door can result. I If, instead, there is a component of the force that is perpendicular to that radial line, a rotation is possible. I If we apply the same force closer to the hinges, it results in less rotation per unit time than if we apply the same force farther from the hinges (this is why door handles are most effective when placed as far as possible from the hinges)

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 16 Rotational Energy and Force Concepts Using Forces to Rotate Bodies about an Axis We’ve been talking about spinning up objects so that they rotate about some axis (some fixed line in space), but we’ve avoided talking about how we get them spinning at some angular speed in the first place. I We know from Newton’s Second Law of Motion that if we want to take an object from a state of motion (e.g. zero angular speed) to a different state of motion (e.g. non-zero angular speed), we must apply a force. I Forces, applied to bodies that can rotate about an axis, that cause a rotation to occur result in what are known as torques. They are the rotational analog of linear force. I We know a few things about torques from our experiences with opening doors: I If the force is applied parallel to radial line that points back to the axis of rotation (either toward the hinges of the door or away from the hinges of the door), then no rotation of the door can result. I If, instead, there is a component of the force that is perpendicular to that radial line, a rotation is possible. I If we apply the same force closer to the hinges, it results in less rotation per unit time than if we apply the same force farther from the hinges (this is why door handles are most effective when placed as far as possible from the hinges)

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 16 Rotational Energy and Force Concepts Using Forces to Rotate Bodies about an Axis We’ve been talking about spinning up objects so that they rotate about some axis (some fixed line in space), but we’ve avoided talking about how we get them spinning at some angular speed in the first place. I We know from Newton’s Second Law of Motion that if we want to take an object from a state of motion (e.g. zero angular speed) to a different state of motion (e.g. non-zero angular speed), we must apply a force. I Forces, applied to bodies that can rotate about an axis, that cause a rotation to occur result in what are known as torques. They are the rotational analog of linear force. I We know a few things about torques from our experiences with opening doors: I If the force is applied parallel to radial line that points back to the axis of rotation (either toward the hinges of the door or away from the hinges of the door), then no rotation of the door can result. I If, instead, there is a component of the force that is perpendicular to that radial line, a rotation is possible. I If we apply the same force closer to the hinges, it results in less rotation per unit time than if we apply the same force farther from the hinges (this is why door handles are most effective when placed as far as possible from the hinges)

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 16 Rotational Energy and Force Concepts Using Forces to Rotate Bodies about an Axis We’ve been talking about spinning up objects so that they rotate about some axis (some fixed line in space), but we’ve avoided talking about how we get them spinning at some angular speed in the first place. I We know from Newton’s Second Law of Motion that if we want to take an object from a state of motion (e.g. zero angular speed) to a different state of motion (e.g. non-zero angular speed), we must apply a force. I Forces, applied to bodies that can rotate about an axis, that cause a rotation to occur result in what are known as torques. They are the rotational analog of linear force. I We know a few things about torques from our experiences with opening doors: I If the force is applied parallel to radial line that points back to the axis of rotation (either toward the hinges of the door or away from the hinges of the door), then no rotation of the door can result. I If, instead, there is a component of the force that is perpendicular to that radial line, a rotation is possible. I If we apply the same force closer to the hinges, it results in less rotation per unit time than if we apply the same force farther from the hinges (this is why door handles are most effective when placed as far as possible from the hinges)

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 16 Rotational Energy and Force Concepts Using Forces to Rotate Bodies about an Axis We’ve been talking about spinning up objects so that they rotate about some axis (some fixed line in space), but we’ve avoided talking about how we get them spinning at some angular speed in the first place. I We know from Newton’s Second Law of Motion that if we want to take an object from a state of motion (e.g. zero angular speed) to a different state of motion (e.g. non-zero angular speed), we must apply a force. I Forces, applied to bodies that can rotate about an axis, that cause a rotation to occur result in what are known as torques. They are the rotational analog of linear force. I We know a few things about torques from our experiences with opening doors: I If the force is applied parallel to radial line that points back to the axis of rotation (either toward the hinges of the door or away from the hinges of the door), then no rotation of the door can result. I If, instead, there is a component of the force that is perpendicular to that radial line, a rotation is possible. I If we apply the same force closer to the hinges, it results in less rotation per unit time than if we apply the same force farther from the hinges (this is why door handles are most effective when placed as far as possible from the hinges)

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 16 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip y(cm) of ~r. This represents the physical “door opening” 1 situation we were just describing. The component of the force parallel to ~r, F~ , does nothing to help rotate ~r || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip ~ y(cm) F~ of r. This represents the physical “door opening” 1 situation we were just describing. The component of the force parallel to ~r, F~ , does nothing to help rotate ~r || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip ~ y(cm) F~ of r. This represents the physical “door opening” 1 situation we were just describing. The component of ~ the force parallel to ~r, F~ , does nothing to help rotate ~r F|| || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip y(cm) ~ of ~r. This represents the physical “door opening” F⊥ F~ 1 situation we were just describing. The component of ~ the force parallel to ~r, F~ , does nothing to help rotate ~r F|| || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip y(cm) ~ of ~r. This represents the physical “door opening” F⊥ F~ 1 situation we were just describing. The component of ~ the force parallel to ~r, F~ , does nothing to help rotate ~r F|| || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip y(cm) ~ of ~r. This represents the physical “door opening” F⊥ F~ 1 situation we were just describing. The component of ~ the force parallel to ~r, F~ , does nothing to help rotate ~r F|| || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip y(cm) ~ of ~r. This represents the physical “door opening” F⊥ F~ 1 situation we were just describing. The component of ~ the force parallel to ~r, F~ , does nothing to help rotate ~r F|| || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque Begin by representing the door as a line on the 3 x-axis, and the vector that points from the hinges (axis of rotation) to the place where a force will be 2 applied as ~r. Then add the force, F~ , applied at the tip y(cm) ~ of ~r. This represents the physical “door opening” F⊥ F~ 1 situation we were just describing. The component of ~ F~ the force parallel to ~r, F||, does nothing to help rotate ~r θ || 0 the door. The component perpendicular to ~r, F~ , can 0 1 2 3 ⊥ start the door rotating about its hinges. It is only this x(cm) perpendicular component that results in a torque. Mathematically, we can represent this as:

τ = rF⊥ = rF sin θ

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 17 Rotational Energy and Force Concepts Representing Torque

3 τ = rF sin θ The above might feel a little familiar. In fact, it looks 2 just like the magnitude of a vector cross product, e.g. y(cm) ~ |~a × ~b| = ab sin θ, where θ is again the opening angle F⊥ F~ 1 between the two vectors. Indeed, torque can be ~ exactly represented in both direction and magnitude ~r F|| 0 θ by a cross-product, 0 1 2 3 x(cm) ~τ = ~r × F~ whose magnitude is rF sin θ and whose direction can be determined by applying the right-hand rule for the vector product (shown left by the circle with the black dot in the middle).

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 18 Rotational Energy and Force Concepts Representing Torque

3 τ = rF sin θ The above might feel a little familiar. In fact, it looks 2 just like the magnitude of a vector cross product, e.g. y(cm) ~ |~a × ~b| = ab sin θ, where θ is again the opening angle F⊥ F~ 1 between the two vectors. Indeed, torque can be ~ exactly represented in both direction and magnitude ~r F|| 0 θ by a cross-product, 0 1 2 3 x(cm) ~τ = ~r × F~ whose magnitude is rF sin θ and whose direction can be determined by applying the right-hand rule for the vector product (shown left by the circle with the black dot in the middle).

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 18 Rotational Energy and Force Concepts Representing Torque

3 τ = rF sin θ The above might feel a little familiar. In fact, it looks 2 just like the magnitude of a vector cross product, e.g. y(cm) ~ |~a × ~b| = ab sin θ, where θ is again the opening angle F⊥ F~ 1 between the two vectors. Indeed, torque can be ~ exactly represented in both direction and magnitude ~r F|| 0 θ by a cross-product, 0 1 2 3 x(cm) ~τ = ~r × F~ whose magnitude is rF sin θ and whose direction can be determined by applying the right-hand rule for the vector product (shown left by the circle with the black dot in the middle).

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 18 Rotational Energy and Force Concepts Representing Torque

3 τ = rF sin θ The above might feel a little familiar. In fact, it looks 2 just like the magnitude of a vector cross product, e.g. y(cm) ~ |~a × ~b| = ab sin θ, where θ is again the opening angle F⊥ F~ 1 between the two vectors. Indeed, torque can be ~ exactly represented in both direction and magnitude ~r F|| 0 θ by a cross-product, 0 1 2 3 x(cm) ~τ = ~r × F~ whose magnitude is rF sin θ and whose direction can be determined by applying the right-hand rule for the vector product (shown left by the circle with the black dot in the middle).

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 18 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Newton’s Second Law of Motion in Rotational Motion

Now that we have seen how forces and rotation go together, can we arrive at a simple statement about how forces, generating torques, then lead to angular accelerations? Indeed, let us begin by considering a point of mass, m, that can rotate about an axis a distance r from that axis. A force, F⊥, is applied to the mass (perpendicular to ~r) and begins to cause an acceleration of the mass tangent to the circular path of motion, at . Then

F⊥ = mat = m(αr)

Note that if we multiply both sides of the above equation by r, we can get torque into this equation:

2 2 I=mr F⊥r = mαr −−−−→ τ = Iα

This is Newton’s Second Law of Motion for a Body that can Rotate, expressing how a torque generates an angular acceleration.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 19 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

If a body is acted upon by a torque, and this results in a change in the state of motion, this must also result in a change in kinetic energy of the body. We already know how to write kinetic energy for a rotating body, and so the work done on the body and the change in its kinetic energy must be related as before:

1 1 W = ∆K −→ W = Iω2 − Iω2 2 f 2 i But how do we go from torque to work? Work is still defined as Z W = F~ · d~s

~ ~ where s is some little bit of path over which the force, F, acts. In rotation, only F⊥ causes the rotation to begin. What is the path over which it acts? Well, as the body begins to ~ rotate (accelerate), F⊥ is acting over a bit of arclength of the circular path, ds = rdθ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 20 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

If a body is acted upon by a torque, and this results in a change in the state of motion, this must also result in a change in kinetic energy of the body. We already know how to write kinetic energy for a rotating body, and so the work done on the body and the change in its kinetic energy must be related as before:

1 1 W = ∆K −→ W = Iω2 − Iω2 2 f 2 i But how do we go from torque to work? Work is still defined as Z W = F~ · d~s

~ ~ where s is some little bit of path over which the force, F, acts. In rotation, only F⊥ causes the rotation to begin. What is the path over which it acts? Well, as the body begins to ~ rotate (accelerate), F⊥ is acting over a bit of arclength of the circular path, ds = rdθ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 20 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

If a body is acted upon by a torque, and this results in a change in the state of motion, this must also result in a change in kinetic energy of the body. We already know how to write kinetic energy for a rotating body, and so the work done on the body and the change in its kinetic energy must be related as before:

1 1 W = ∆K −→ W = Iω2 − Iω2 2 f 2 i But how do we go from torque to work? Work is still defined as Z W = F~ · d~s

~ ~ where s is some little bit of path over which the force, F, acts. In rotation, only F⊥ causes the rotation to begin. What is the path over which it acts? Well, as the body begins to ~ rotate (accelerate), F⊥ is acting over a bit of arclength of the circular path, ds = rdθ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 20 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

If a body is acted upon by a torque, and this results in a change in the state of motion, this must also result in a change in kinetic energy of the body. We already know how to write kinetic energy for a rotating body, and so the work done on the body and the change in its kinetic energy must be related as before:

1 1 W = ∆K −→ W = Iω2 − Iω2 2 f 2 i But how do we go from torque to work? Work is still defined as Z W = F~ · d~s

~ ~ where s is some little bit of path over which the force, F, acts. In rotation, only F⊥ causes the rotation to begin. What is the path over which it acts? Well, as the body begins to ~ rotate (accelerate), F⊥ is acting over a bit of arclength of the circular path, ds = rdθ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 20 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

If a body is acted upon by a torque, and this results in a change in the state of motion, this must also result in a change in kinetic energy of the body. We already know how to write kinetic energy for a rotating body, and so the work done on the body and the change in its kinetic energy must be related as before:

1 1 W = ∆K −→ W = Iω2 − Iω2 2 f 2 i But how do we go from torque to work? Work is still defined as Z W = F~ · d~s

~ ~ where s is some little bit of path over which the force, F, acts. In rotation, only F⊥ causes the rotation to begin. What is the path over which it acts? Well, as the body begins to ~ rotate (accelerate), F⊥ is acting over a bit of arclength of the circular path, ds = rdθ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 20 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

If a body is acted upon by a torque, and this results in a change in the state of motion, this must also result in a change in kinetic energy of the body. We already know how to write kinetic energy for a rotating body, and so the work done on the body and the change in its kinetic energy must be related as before:

1 1 W = ∆K −→ W = Iω2 − Iω2 2 f 2 i But how do we go from torque to work? Work is still defined as Z W = F~ · d~s

~ ~ where s is some little bit of path over which the force, F, acts. In rotation, only F⊥ causes the rotation to begin. What is the path over which it acts? Well, as the body begins to ~ rotate (accelerate), F⊥ is acting over a bit of arclength of the circular path, ds = rdθ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 20 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

If a body is acted upon by a torque, and this results in a change in the state of motion, this must also result in a change in kinetic energy of the body. We already know how to write kinetic energy for a rotating body, and so the work done on the body and the change in its kinetic energy must be related as before:

1 1 W = ∆K −→ W = Iω2 − Iω2 2 f 2 i But how do we go from torque to work? Work is still defined as Z W = F~ · d~s

~ ~ where s is some little bit of path over which the force, F, acts. In rotation, only F⊥ causes the rotation to begin. What is the path over which it acts? Well, as the body begins to ~ rotate (accelerate), F⊥ is acting over a bit of arclength of the circular path, ds = rdθ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 20 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

Z W = F~ · d~s

If we utilize this information in the definition of work, we arrive at the following: Z Z ~ W = F · d~s = F⊥rdθ

Thus we arrive at the relationship to torque:

Z θf W = τdθ θi

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 21 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

Z W = F~ · d~s

If we utilize this information in the definition of work, we arrive at the following: Z Z ~ W = F · d~s = F⊥rdθ

Thus we arrive at the relationship to torque:

Z θf W = τdθ θi

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 21 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

Z W = F~ · d~s

If we utilize this information in the definition of work, we arrive at the following: Z Z ~ W = F · d~s = F⊥rdθ

Thus we arrive at the relationship to torque:

Z θf W = τdθ θi

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 21 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

Z W = F~ · d~s

If we utilize this information in the definition of work, we arrive at the following: Z Z ~ W = F · d~s = F⊥rdθ

Thus we arrive at the relationship to torque:

Z θf W = τdθ θi

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 21 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

Z W = F~ · d~s

If we utilize this information in the definition of work, we arrive at the following: Z Z ~ W = F · d~s = F⊥rdθ

Thus we arrive at the relationship to torque:

Z θf W = τdθ θi

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 21 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

Z W = F~ · d~s

If we utilize this information in the definition of work, we arrive at the following: Z Z ~ W = F · d~s = F⊥rdθ

Thus we arrive at the relationship to torque:

Z θf W = τdθ θi

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 21 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

And so by the Work-Kinetic Energy Theorem, the following must be true for a body acted on by a series of torques and experiencing a change in the state of rotational motion:

Z θf 1 2 1 2 Iωf − Iωi = τdθ 2 2 θi If we consider just a little bit of work done by the torque over some little change in angular location in the motion, dW = τdθ, we can then think about the energy per unit time (power) that is being supplied to drive (or slow) the rotation:

dW dθ P = = τ = τω dt dt

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 22 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

And so by the Work-Kinetic Energy Theorem, the following must be true for a body acted on by a series of torques and experiencing a change in the state of rotational motion:

Z θf 1 2 1 2 Iωf − Iωi = τdθ 2 2 θi If we consider just a little bit of work done by the torque over some little change in angular location in the motion, dW = τdθ, we can then think about the energy per unit time (power) that is being supplied to drive (or slow) the rotation:

dW dθ P = = τ = τω dt dt

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 22 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

And so by the Work-Kinetic Energy Theorem, the following must be true for a body acted on by a series of torques and experiencing a change in the state of rotational motion:

Z θf 1 2 1 2 Iωf − Iωi = τdθ 2 2 θi If we consider just a little bit of work done by the torque over some little change in angular location in the motion, dW = τdθ, we can then think about the energy per unit time (power) that is being supplied to drive (or slow) the rotation:

dW dθ P = = τ = τω dt dt

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 22 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

And so by the Work-Kinetic Energy Theorem, the following must be true for a body acted on by a series of torques and experiencing a change in the state of rotational motion:

Z θf 1 2 1 2 Iωf − Iωi = τdθ 2 2 θi If we consider just a little bit of work done by the torque over some little change in angular location in the motion, dW = τdθ, we can then think about the energy per unit time (power) that is being supplied to drive (or slow) the rotation:

dW dθ P = = τ = τω dt dt

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 22 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

And so by the Work-Kinetic Energy Theorem, the following must be true for a body acted on by a series of torques and experiencing a change in the state of rotational motion:

Z θf 1 2 1 2 Iωf − Iωi = τdθ 2 2 θi If we consider just a little bit of work done by the torque over some little change in angular location in the motion, dW = τdθ, we can then think about the energy per unit time (power) that is being supplied to drive (or slow) the rotation:

dW dθ P = = τ = τω dt dt

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 22 Rotational Energy and Force Concepts Revisiting the Work-Kinetic Energy Theorem

And so by the Work-Kinetic Energy Theorem, the following must be true for a body acted on by a series of torques and experiencing a change in the state of rotational motion:

Z θf 1 2 1 2 Iωf − Iωi = τdθ 2 2 θi If we consider just a little bit of work done by the torque over some little change in angular location in the motion, dW = τdθ, we can then think about the energy per unit time (power) that is being supplied to drive (or slow) the rotation:

dW dθ P = = τ = τω dt dt

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 22 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Taking Stock Again: Angular Motion and Linear Motion

LINEAR QUANTITY (1-D) ANGULAR QUANTITY

F = ma τ = Iα 1 1 K = mv 2 K = Iω2 2 2 Z Z W = Fdx W = τdθ P = Fv P = τω

The above are just some examples. In principle, all of the 1-D equations of energy and force have an angular analog. You can obtain one from the other by changing x → θ, v → ω, a → α, m → I, and F → τ.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 23 Rotational Energy and Force Concepts Key Ideas

The key ideas that we have explored in this section of the course are as follows: I We began to think about energy and force concepts as they related to extended, rigid, rotating bodies. I We saw that there are rotational analogs to linear concepts, like

kinetic energy. “Sprunger Table Saw blade” by I We have seen how forces result in rotations, and explored the lungstruck, available under Creative concept of a torque. Commons from Flickr I We have connected the force and energy concepts through work, and completed a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 24 Rotational Energy and Force Concepts Key Ideas

The key ideas that we have explored in this section of the course are as follows: I We began to think about energy and force concepts as they related to extended, rigid, rotating bodies. I We saw that there are rotational analogs to linear concepts, like

kinetic energy. “Sprunger Table Saw blade” by I We have seen how forces result in rotations, and explored the lungstruck, available under Creative concept of a torque. Commons from Flickr I We have connected the force and energy concepts through work, and completed a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 24 Rotational Energy and Force Concepts Key Ideas

The key ideas that we have explored in this section of the course are as follows: I We began to think about energy and force concepts as they related to extended, rigid, rotating bodies. I We saw that there are rotational analogs to linear concepts, like

kinetic energy. “Sprunger Table Saw blade” by I We have seen how forces result in rotations, and explored the lungstruck, available under Creative concept of a torque. Commons from Flickr I We have connected the force and energy concepts through work, and completed a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 24 Rotational Energy and Force Concepts Key Ideas

The key ideas that we have explored in this section of the course are as follows: I We began to think about energy and force concepts as they related to extended, rigid, rotating bodies. I We saw that there are rotational analogs to linear concepts, like

kinetic energy. “Sprunger Table Saw blade” by I We have seen how forces result in rotations, and explored the lungstruck, available under Creative concept of a torque. Commons from Flickr I We have connected the force and energy concepts through work, and completed a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 24 Rotational Energy and Force Concepts Key Ideas

The key ideas that we have explored in this section of the course are as follows: I We began to think about energy and force concepts as they related to extended, rigid, rotating bodies. I We saw that there are rotational analogs to linear concepts, like

kinetic energy. “Sprunger Table Saw blade” by I We have seen how forces result in rotations, and explored the lungstruck, available under Creative concept of a torque. Commons from Flickr I We have connected the force and energy concepts through work, and completed a set of rotational analogs to linear energy and force concepts.

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 24 References References I

S. Sekula (SMU) SMU — PHYS 1303 SPRING, 2019 25