Lecture 19 Physics I Chapter 12

Center of Mass Moment of Inertia

Course website: http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics IN THIS CHAPTER, you will start discussing rotational dynamics

Today we are going to discuss:

Chapter 12:

 Rotation about the Center of Mass: Section 12.2

(skip “Finding the CM by Integration)  Rotational Kinetic Energy: Section 12.3  Moment of Inertia: Section 12.4

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Center of Mass (CM)

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Center of Mass (CM) idea

We know how to address these problems: How to describe motions like these? It is a rigid object.

Translational plus rotational motion

We also know how to address this motion of a single particle - kinematic equations

The general motion of an object can be considered as the sum of translational motion of a certain point, plus rotational motion about that point.

That point is called the center of mass point.

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Center of Mass: Definition

    m1r1  m2r2  m3r3 Position vector of the CM: r  M  m1  m2  m3 CM total mass of the system m1  m2  m3

n  1  The center of mass is the rCM   miri mass-weighted center of M i1 the object Component form: m2 1 n m1  xCM  mi xi r2  M i1 rCM  (xCM , yCM , zCM )  n r1 1 yCM  mi yi M i1

r 1 n 3 m3 zCM  mi zi M i1

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Example Center of Mass Position (2 particles)

What is the center of mass of 2 point masses (mA=1 kg and mB=3 kg), at two different points: A=(0,0) and B=(2,4)?

m x  m x By definition: x  A A B B CM (m  m ) A B B=(2,4) mA yA  mB yB m =3kg yCM  B (mA  mB )

mA=1 kg and mB=3 kg CM r (1  0) (3 2) CM xCM  1.5 1 3 A=(0,0) mA=1kg (1  0) (3 4) y   3 CM 1 3 Or in a vector form:  ˆ ˆ rCM 1.5i  3j

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics CM of a solid object (But we are not going to use this since I promised to avoid integration) Let’s find CM of an extended body: n  1  Before, for many particles we had rCM   miri M i1 Now, let’s divide extended mass into smaller sections ∆m m i i 1 r  m r  CM M  i i ri i  1  rCM  lim miri m 0  M i i

1 r  rdm CM M 

1 1 1 xCM   xdm yCM  ydm zCM   zdm M M  M

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics CM of solid symmetrical objects y The easiest trick is to use symmetry coordinates (-5;4) ... m(5)  m(5) ... So, contributions to CM from m xCM  M symmetrical points cancel CM x ... m(4)  m(4) ... each other and, as a result, yCM  the CM coordinates are (0;0) m M (5;-4) coordinates

CM CM

If we break a symmetry, the CM will be shifted

CM CM Old CM Old CM

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics ConcepTest Center of Mass

A) higher The disk shown below in (1) clearly has its center of mass at the center. B) lower Suppose a smaller disk is cut out as shown in (2). C) at the same place Where is the center of mass of (2) as compared to (1) ? D) there is no definable CM in this case

(1) (2)

CM X CM Rotational Kinetic Energy and Moment of Inertia

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Rotational Kinetic Energy

Consider a pure rotation of a solid object. A rotating object has kinetic energy because all particles in the object are in motion. Now, let’s divide extended mass into smaller sections mi  1 2 Krot   2 mivi vi   ri 2 K  1 m ( r ) 1 2 2 rot  2 i i   2 mi ri 1 2 2 Krot  2   miri v  i We ended up with a new structure. Let’s give it a nice name and a symbol: ri m Moment of inertia 2 i I miri

Axis of Rotation Rotational Kinetic Energy 1 2 The units of moment of (The kinetic energy due to rotation Krot  2 I inertia are kg m2. is called rotational kinetic energy.) 1 K  mv 2 Translational Kinetic Energy trans 2 Moment of inertia is the rotational equivalent of mass. m I v  This is not a new form of energy, merely the familiar kinetic energy of motion written in a new way. PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Why moment of inertia is needed to describe rotation?

Let’s try to move objects How about trying to rotate these kids up? translationally? m m r M>m m We know that it is easier to move a lighter object than heavier. R>r So information about mass is enough to describe the “translational challenge”. We know that it is easier to rotate the same object when it is closer to an axis of rotation. So information about mass is NOT enough to describe the “rotational challenge”. • So it should depend on r and m. That is why the moment of inertia appeared in the “rotational game” , I=mr2 Mass farther from the rotation axis contributes more to the moment of inertia than mass nearer the axis. Moment of inertia is the rotational equivalent of mass.

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics ConcepTest Center of Mass  Which dumbbell has the larger moment of inertia about the midpoint of the rod? The connecting rod is massless. A) Dumbbell A. B) Dumbbell B. C) Their moments of inertia are the same.

Axis of rotation

2 2 2 I  miri  m1r1  m2r2 Distance from the axis is more important than mass. 2 2 2 2 I A  m(R / 2)  m(R / 2)  2mR / 4  mR / 2 2 2 2 2 IB  (m/ 2)(R)  (m/ 2)(R)  2mR / 2  mR Moments of Inertia for many solid objects

As we did for center of mass, divide a solid object into many small cells of mass Δm and let Δ m → 0. The moment of inertia sum becomes

where r is the distance from the rotation axis.

The procedure is much like calculating the center of mass. One rarely needs to do this integral because moments of inertia of common shapes are tabulated.

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Parallel Axis Theorem (w/out proof)

2 Moment of inertia depends on the axis of rotation. I  miri How can we relate the moment of inertia measured relative to different axis of rotation?

The moment of inertia about any axis parallel to that axis through the center of mass is given by I CM I 2 I  ICM  Md d CM I : moment of inertia about any parallel axis

ICM : moment of inertia about an axis through its center of mass M : total mass d : distance from a parallel axis to the center of mass.

Relative to which axis is the moment of inertia the smallest? BTW: The moment of inertia of any object about an axis through its center of mass is the minimum moment of inertia for an axis in that direction in space.

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Example Parallel Axis Theorem: Sphere

For a uniform sphere of radius r0

Moment of inertia for the sphere, rotating about an axis 2 2 through its center of mass ICM  5 Mr0

Moment of inertia for the sphere about an axis going through the edge of the sphere?

Apply Parallel Axis Theorem:

2 2 2 2 7 2 I  ICM  Md  5 Mr0  Mr0  5 Mr0

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics CM of Solid Objects (nice trick) How to deal with objects like this? Divide it into symmetrical objects

m1 m1

CM m2 m2 CM of the original object CM

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics Thank you See you on Wednesday

PHYS.1410 Lecture 19 Danylov Department of Physics and Applied Physics