Physics 1 Mechanics – Rotational

We began studying the linear motion of objects by analyzing the forces that cause motion using Newton’s Laws. Later, we later introduced an alternative way to analyze motion through the concept of energy. In a similar fashion, we also began our study of rotational motion using the analyses of forces. As you may have guessed, we can also apply the concept of energy to the study of rotational motion. The rotational energy equations are direct analogues to the translational energy ones. The equations are presented below without proof; however, the direct analogies should be obvious.

Translational Rotational Kinetic Energy 1 1 퐾 = 푚푣2 → 퐾 = 퐼휔2 푡푟푎푛 2 푟표푡 2 done by a force to move an object Work done by a to rotate an object over a distance. through an angle. 푏 → 휃2 푊푡푟푎푛 = ∫ 푭 ∙ 푑풍 푊푟표푡 = ∫ 휏 푑휃 푎 휃1 Translational Rotational Power 푃푡푟푎푛 = 푭 ∙ 풗 → 푃푟표푡 = 휏휔 Translational Work-Energy Principle Rotational Work-Energy Principle 푊푡푟푎푛 = Δ퐾푡푟푎푛 → 푊푟표푡 = Δ퐾푟표푡

Let’s do some examples to see how these relationships can be used to systems that experience rotational motion only.

Example 1: In the first example from the previous lesson, we found the acceleration of a basic pulley system using force techniques, (note we found 푎 = 6.8 푚⁄푠2). In this example we’ll redo the problem using energy techniques. The problem is repeated below for convenience.

A pulley system is arranged as shown below with 푚1 = 20 푘𝑔 and 푚2 = 2 푘𝑔. Find the acceleration of the system assuming the pulley has a of 8 푘𝑔 and radius of 0.25 푚.

푚2

푚1

Solution 1: We’ll use the principal to analyze this example. For this we will choose an initial time to be when the blocks are at rest and separated by a distance, ℎ = 1 푚, and a final time when the blocks are at the same height. For convenience we choose the location that they meet to be our zero gravitational height. The figure below illustrates the scenario.

퐼푛𝑖푡𝑖푎푙 푇𝑖푚푒 퐹𝑖푛푎푙 푇𝑖푚푒

푦

푚1 ℎ⁄2 0 0 -ℎ⁄2

푚2

We are already familiar with the kinetic energy associated with the moving blocks, i.e., linear 2 kinetic energy, 퐾푡푟푎푛 = 1⁄2 푚푣 . For this analysis, since we consider the pulley to have mass, it too has an associated kinetic energy. The motion of the pulley, however, is rotational and 2 therefore, it has rotational kinetic energy, 퐾푟표푡 = 1⁄2 퐼휔 . With this we can write the conservation of energy statement as follows.

퐾푓 + 푈푓 = 퐾푖 + 푈푖 1 1 1 ℎ ℎ 푚 푣 2 + 푚 푣 2 + 퐼휔 2 = 푚 𝑔 + 푚 𝑔 (− ) 2 1 푓 2 2 푓 2 푓 1 2 2 2 Next, we substitute 휔 = 푣⁄푅, and 퐼 = 1⁄2 푀푅2 and solve for the velocity, 푣.

2 1 1 1 1 푣푓 ℎ ℎ 푚 푣 2 + 푚 푣 2 + ( 푀푅2) ( ) = 푚 𝑔 − 푚 𝑔 2 1 푓 2 2 푓 2 2 푅2 1 2 2 2 1 1 1 ℎ 푣 2 ( 푚 + 푚 + 푀) = 𝑔 (푚 − 푚 ) 푓 2 1 2 2 4 2 1 2 2 푣푓 (2푚1 + 2푚2 + 푀) = 2𝑔ℎ(푚1 − 푚2)

2𝑔ℎ(푚1 − 푚2) 푣푓 = √ (2푚1 + 2푚2 + 푀)

2 ∙ 9.8 ∙ 1(20 − 2) 푣 = √ 푓 (2 ∙ 20 + 2 ∙ 2 + 8)

푣푓 = 2.6 푚⁄푠 Next, we use kinematics to find the acceleration.

2 2 2푎∆푦 = 푣푓 + 푣푖 2 2 푣푓 + 푣푖 푎 = 2(ℎ⁄2) 2.62 + 0 푎 = 2 ∙ (1⁄2) 푎 = 6.8 푚⁄푠2

Which is the same answer we obtained using the force based method in the previous lesson!

Example 3: A rod of mass 1 푘𝑔 and length 0.25 푚 is hinged at one end. The rod is held at rest horizontally before being released. Find the of the rod when it reaches the vertical position. Then find the linear speed of the end of the rod and compare this with the speed of an object that was dropped from the same height.

푙

푙 2

Solution 3: We can again use the conservation of energy principal. In this case, the gravitational potential energy of the rod is converted into rotational kinetic energy when the rod is released. For the purposes of computing the gravitational potential energy of the rod we treat it as a point object located at its center of mass. As the figure shows this point on the rod drops a distance of 푙/2 when the rod is vertically oriented. The rotational inertia for a rod about one of its ends is 퐼 = 1⁄3 푚푙2. Finally, using the vertical position as the zero gravitational potential energy position we have

퐾푓 + 푈푓 = 퐾푖 + 푈푖 1 푙 퐼휔2 = 푚𝑔 2 2 1 푚푙2휔2 = 푚𝑔푙 3 휔 = √3𝑔⁄푙

휔 = √3 ∙ 9.8⁄0.25 휔 = 10.84 푟푎푑⁄푠푒푐 The linear speed for the end of the rod is then given as 푣 = 푙휔 3𝑔 푣 = 푙√ 푙 푚 푣 = √3𝑔푙 = 2.71 ⁄푠 The linear speed of a point object that is dropped from the same height, 푙, can also be found from the conservation of energy as follows.

퐾푓 = 푈푖 1 푚푣2 = 푚𝑔푙 2 푚 푣 = √2𝑔푙 = 2.21 ⁄푠

Example 4: A uniform solid disk of radius 9.0 푚 and mass 20 푘𝑔 is free to rotate on a frictionless pivot through a point on its rim as shown.

푃𝑖푣표푡 푅

If the disk is released from rest in the position shown by the solid circle, what is the speed of the center of mass when the disk reaches the position indicated by the dashed circle?

Solution 4: We can use the conservation of energy once more to analyze this system. The center of mass changes its vertical position by 푅 as the disk rotates from its two positions. Using the final position as the zero gravitational potential energy position, the final angular velocity can be found as follows.

퐾푓 + 푈푓 = 퐾푖 + 푈푖 1 퐼휔2 + 0 = 0 + 푚𝑔푅 2 퐼휔2 = 2푚𝑔푅 2푚𝑔푅 휔 = √ 퐼 To find the rotational inertia we need to use the parallel axis theorem since the axis of is not at the center of mass. 1 3 퐼 = 푚푅2 + 푚푅2 = 푚푅2 2 2 Substituting, we have

2푚𝑔푅 휔 = √ 3 2 2 푚푅

4𝑔 휔 = √ 3푅

4 ∙ 9.8 휔 = √ 3 ∙ 9.0

휔 = 1.2 푟푎푑⁄푠푒푐

The velocity of the center of mass is then given as

푣푐푚 = 푅휔 = 9.0 ∙ 1.2 = 10.8 푚⁄푠

Example 5: A solid disk is wrapped with a light string. The disk is held in place about its center of mass while the string is pulled with a constant force of 5.0 푁. The disk has a mass of 5.0 푘𝑔 and radius of 0.25 푚. What is the angular speed of the disk after the disk has been made to rotate through 10 complete revolutions?

퐹

푅

Solution 5: The tension force from the string is doing work on the disk, which is equivalent to the change in rotational kinetic energy of the disk.

푊푟표푡 = ∆퐾푟표푡

푊푟표푡 = 퐾푓 − 퐾푖

In this case, since the disk is initially at rest, we have 1 1 푊 = 퐾 = 퐼휔2 = 푚푅2휔2 푟표푡 푓 2 4 The work can also be expressed using a torque integral as follows.

휃2 푊푟표푡 = ∫ 휏 푑휃 휃1 Since the torque is constant and equal to 휏 = 퐹푅, the integral can be written as

휃2 푊푟표푡 = 퐹푅 ∫ 푑휃 휃1 푊푟표푡 = 퐹푅∆휃 Equating the two we have expressions for work we can solve for the angular speed as follows. 1 푚푅2휔2 = 퐹푅∆휃 4

4퐹∆휃 휔 = √ 푚푅

4 ∙ 5 ∙ (10 ∙ 2휋) 휔 = √ 5 ∙ 0.25

휔 = 31.7 푟푎푑⁄푠푒푐

Final Summary Rotational Energy

Rotational Kinetic Energy 1 퐾 = 퐼휔2 푟표푡 2

Work done by a Torque to Rotate an Object Through an Angle. 휃2 푊푟표푡 = ∫ 휏 푑휃 휃1

Rate of Rotational Work (Rotational Power) 푃푟표푡 = 휏휔

Rotational Work-Energy Principle 푊푟표푡 = Δ퐾푟표푡

By: ferrantetutoring