## Section 9.17 Rotation Bootcamp

### Subsection 9.17.1 Describing Motion

###### Problem 9.17.1. Quantifying Rotation of a Disk.

Follow the link: Checkpoint 9.1.7.

### Subsection 9.17.2 Rotational Speed and Velocity

###### Problem 9.17.2. Angular Velocity from Angle versus Time Plot - Linear Segments.

Follow the link: Checkpoint 9.3.3.

###### Problem 9.17.3. Angular Velocity from Angle versus Time Plot - Curved.

Follow the link: Checkpoint 9.3.5.

### Subsection 9.17.3 Rotational Acceleration

###### Problem 9.17.4. Angular Acceleration from Angular Velocity versus Time Plot - Linear Segments.

Follow the link: Checkpoint 9.4.4.

### Subsection 9.17.4 Constant Rotational Acceleration

###### Problem 9.17.5. Spinning a Centrifuge to Full Speed.

Follow the link: Checkpoint 9.5.2.

### Subsection 9.17.5 Angular Momentum

###### Problem 9.17.6. Magnitude and Direction of Angular Momentum of a Ball Moving in a Circle.

Follow the link: Checkpoint 9.6.4.

###### Problem 9.17.7. Speed of a Ball Moving in a Circle with Angular Momentum Given.

Follow the link: Checkpoint 9.6.6.

### Subsection 9.17.6 Moment of Inertia

###### Problem 9.17.8. Moment of Inertia of a Baton about an Irregular Axis.

Follow the link: Checkpoint 9.7.3.

###### Problem 9.17.9. Moment of Inertia and Angular Momentum of a Three Particle System.

Follow the link: Checkpoint 9.7.6.

###### Problem 9.17.10. Moment of Inertia and Angular Momentum of a Three Particle System Different Axis.

Follow the link: Checkpoint 9.7.9.

###### Problem 9.17.11. A Chain of Rings Rotating About an Axis Through One End.

Follow the link: Checkpoint 9.7.13.

###### Problem 9.17.12. Moment of Inertia of a Disk with a Hole in the Center - Negative Mass Trick.

Follow the link: Checkpoint 9.7.17.

###### Problem 9.17.13. Moment of Inertia of a Disk with Off-Axis Hole - Negative Mass Trick.

Follow the link: Checkpoint 9.7.19.

###### Problem 9.17.14. (Calculus) Moment of Inertia of Rod about Axis Through the Center TODO.

Follow the link: Checkpoint 9.7.23.

###### Problem 9.17.15. (Calculus) Moment of Inertia of Triangular Plate about an Axis Through the Corner.

Follow the link: Checkpoint 9.7.25.

###### Problem 9.17.16. (Calculus) Moment of Inertia of Right Triangular Plate About one Edge.

Follow the link: Checkpoint 9.7.27.

###### Problem 9.17.17. (Calculus) Moment of Inertia of Thin Disk about an Axis Through the Center and Perpendicular to Disk TODO.

Follow the link: Checkpoint 9.7.29.

###### Problem 9.17.18. (Calculus) Moment of Inertia of Thin Disk about an Axis Through the Center and Coplanar to Disk TODO.

Follow the link: Checkpoint 9.7.31.

###### Problem 9.17.19. (Calculus) Moment of Inertia of a Sphere about an Axis Through the Center TODO.

Follow the link: Checkpoint 9.7.33.

### Subsection 9.17.7 Torque

###### Problem 9.17.20. Torque on a Bar by a Force Acting at an Angle.

Follow the link: Checkpoint 9.8.8.

###### Problem 9.17.21. Torque of Particles Hitting a Flywheel.

Follow the link: Checkpoint 9.8.11.

###### Problem 9.17.22. Computing Net Torque on a Beam.

Follow the link: Checkpoint 9.8.12.

### Subsection 9.17.8 Rotational Second Law

###### Problem 9.17.23. Angular Acceleration of a Pulley.

Follow the link: Checkpoint 9.9.1.

###### Problem 9.17.24. Coupled Motion of a Block Attached to an Unwinding Tape on a Rotating Wheel.

Follow the link: Checkpoint 9.9.4.

###### Problem 9.17.25. The Atwood Machine.

Follow the link: Checkpoint 9.9.8.

### Subsection 9.17.9 Rotational Impulse

###### Problem 9.17.26. Rotating a Pulley by a Sudden Pull on a Rope Wound Around the Pulley.

Follow the link: Checkpoint 9.10.4.

### Subsection 9.17.10 Rotational Impulse and Angular Momentum

###### Problem 9.17.27. Rotational Impulse Causing Change in Angular Momentum.

Follow the link: Checkpoint 9.11.3.

### Subsection 9.17.11 Conservation of Angular Momentum

###### Problem 9.17.28. Increase in Spin Rate when a Star Collapses to a Neutron Star.

Follow the link: Checkpoint 9.12.3.

###### Problem 9.17.29. Walking on a Rotating Platform.

Follow the link: Checkpoint 9.12.4.

### Subsection 9.17.12 Collision and Angular Momentum

###### Problem 9.17.30. A Ball Striking a Rotating Wheel and Bouncing Back.

Follow the link: Checkpoint 9.13.1.

### Subsection 9.17.13 Rotational Work

###### Problem 9.17.31. Work in Tightening a Bolt.

Follow the link: Checkpoint 9.14.2.

### Subsection 9.17.14 Rotational Kinetic Energy

###### Problem 9.17.32. Kinetic Energy of a Baton Rotating about the Center.

Follow the link: Checkpoint 9.15.1.

###### Problem 9.17.33. Kinetic Energy of a Baton Rotating about a Point away from the Center.

Follow the link: Checkpoint 9.15.2.

###### Problem 9.17.34. Work to Change Rotational Kinetic Energy of a Disk.

Follow the link: Checkpoint 9.15.3.

### Subsection 9.17.15 Rolling Motion

###### Problem 9.17.35. Penny Rolled in a Circular Groove.

Follow the link: Checkpoint 9.16.2.

###### Problem 9.17.36. Kinetic Energy of a Rolling Bowling Ball.

Follow the link: Checkpoint 9.16.4.

###### Problem 9.17.37. Speed of a Ball Rolling Down an Incline.

Follow the link: Checkpoint 9.16.5.

###### Problem 9.17.38. Acceleration of Rolling Drum Down an Incline.

Follow the link: Checkpoint 9.16.7.

###### Problem 9.17.39. Rolling Motion of Bicyle.

Follow the link: Checkpoint 9.16.9.

###### Problem 9.17.40. Acceleration of a Yoyo Falling with Thread Unwinding.

Follow the link: Checkpoint 9.16.12.

###### Problem 9.17.41. Spherical Ball Rolling Down a Straight Track.

Follow the link: Checkpoint 9.16.15.

###### Problem 9.17.42. Pushing a Heavy Drum Up an Incline in Rolling Motion.

Follow the link: Checkpoint 9.16.16.

### Subsection 9.17.16 Miscellaneous

###### Problem 9.17.43. Unwinding Rope on a Wheel.

A wheel of mass \(M\) and radius \(R\) is mounted vertically on an axle of radius \(r\text{.}\) A light rope is wound on the wheel. When the rope is pulled, it unwinds at a distance \(\dfrac{2}{3} R\) from the center of the wheel without slipping as shown.

(a) Suppose the tension in the rope has a magnitude \(T\) at some instant in time, what will be the angular acceleration of the wheel at that instant if the effects of friction at the axle can be neglected?

(b) Suppose you find that a tension of magnitude \(T_0\) is needed to pull the string at a constant speed, what friction must be acting at the axle if friction acts at a distance \(r\) from the center of the wheel?

(a) The lever arm of \(T\) is \(\dfrac{2}{3} R\text{.}\)

(b) The torques are balanced.

(a) \(\dfrac{4}{3}\, \dfrac{T}{MR}\text{,}\) (b) \(\dfrac{2}{3}\, \dfrac{R}{r}\, T_0\text{.}\)

(a) We recognize that the leverarm of \(T\) is \(\dfrac{2}{3} R\text{.}\) This gives the following for the rotational second law of motion.

where the moment of inertia can be approximated by the moment of inertia of a disk.

Therefore magnitude of the angular acceleration is

The direction of the angular acceleration will be to to the right in the figure, i.e., counterclockwise sense when seen from the right.

(b) Now, the two torques are balanced.

Therefore,

###### Problem 9.17.45. Angular Acceleration of a Falling Plank Hinged at One End.

A plank of mass \(M\) and length \(L\) is hinged at one end. Initially, the plank is supported at the free end so that the plank is horizontal. When the plank is released, the imbalance of the torque on the plank causes a varying angular acceleration of the plank.

(a) Find the angular acceleration of the plank at two instants immediately after the plank is released.

(b) Find the angular acceleration of the plank when the plank makes an angle \(\theta\) with the horizontal.

(a) Weight acts at the center, (b) Figure out the lever arm of weight when the plank makes an angle \(\theta\text{.}\)

(a) \(\dfrac{3}{2}\, \dfrac{g}{L}\text{,}\) (b) \(\dfrac{3}{2}\, \dfrac{g}{L}\, \sin\,\theta\text{.}\)

(a) Only weight acts at points away from the pivot. We can place the entire weight at the center of the plank. This gives the torque by weight \(MgL/2\text{.}\) Since the plank rotates about an axis through one end, we wil luse the formula of moment of inertia for a rod with axis at the end. These give

Therefore,

(b) Here, the lever arm changes to

while the moment of inertial is same. Therefore, we get

Therefore,

###### Problem 9.17.49. (Calculus) Meter Stick Sliding on a Table from Vertical to Horizontal Orientation.

A meter-stick is set almost vertically on a frictionless table. When the stick is let go at rest from the nearly vertical orientation it falls to the table. Since there is no horizontal force on the stick and it starts out at rest, the center of mass of the stick falls straight down. Find the speed of the CM as a function of the position of the CM with respect to the table. Hint:

Let the \(y\)-axis be the vertical direction and the \(z\)-axis into-the-page, and write the \(y\)-component of equation of motion for the translation of CM and the \(z\)-component of the rotational motion of the stick about the CM.

Numerical: when it has fallen a vertical distance of \(10\text{ cm}\text{,}\) answer is \(1.01\text{ m/s}\text{.}\)

Use energy conservation with energy when vertical to when at an angle.

\(- \left[\frac{3g l (1-\cos\theta)}{3+ \textrm{cosec}^2\theta} \right]^{1/2}\ \hat j\text{.}\)

Since there is no horizontal force on the stick and since the CM of the stick does not have any initial velocity in the horizontal direction, the CM will fall vertically down. As the stick falls down, the lower part of the stick will move one way and the upper part the other way. The rotational motion of the stick about an axis through the CM is possible due to the normal force from the table. For concreteness we will point the positive \(y\)-axis up and measure the angle of rotation with respect to the vertical as shown in Figure 9.17.51.

The relation between the \(y\)-coordinate of the CM and the angle \(\theta\) is

The time derivative \(y\) will give the \(y\)-component velocity and the time derivative of \(\theta\) will give the \(z\)-component of the angular velocity.

Here the choice of the angle gives \(d\theta/dt\) positive since \(\theta\) will increase with time. As the \(y\)-coordinate of the CM is decreasing with time we expect the \(y\)-component of the velocity to be negative, i.e. the velocity pointed towards the negative \(y\)-axis.

Now, we use the energy conservation between the initial time \(t=0\) and at an arbitrary time \(t=t\text{.}\) The reference of the potential energy will be take to be zero when the CM of the stick is on the table. The CM is initially at \(y=l/2\) and the stick is at rest. The CM at an arbitrary time is at \(y=y\text{,}\) and the CM is translating with speed \(|dy/dt|\) and the stick is rotating about an axis through the CM parallel to the \(z\)-axis in the figure with the angular speed \(|d\theta/dt|\text{.}\) Therefore,

where

Now, using Eq. (9.17.1) we can write the speeds in terms of \(|dy/dt|\) and solve for \(|dy/dt|\text{.}\)

Therefore, the velocity of the CM is

where I have used the negative root in Eq. (9.17.2) to agree with the motion down as shown in the figure.

###### Problem 9.17.52. Tape Unwounding on Two Disks with Tape Going Over a Pulley.

A “massless” tape is wound on one disk of mass \(m_1\) and radius \(R_1\text{.}\) The tape then goes over a pulley of mass \(M\) and radius \(R\) and the other end of the tape is wound on another disk of mass \(m_2\) and radius \(R_2\) as shown in Figure 9.17.53. The disks are then released from rest with tapes taut on each side. Assume the tape unwinds smoothly on each side. Determine the accelerations of each mass and the angular acceleration of the pulley.

Set up equations of motion for the three objects alongwith connections between their accelerations.

Acceleration of pulley, \(\alpha = \dfrac{|m_1-m_2|}{m_1 + m_2 + \frac{3}{2}M}\ \dfrac{g}{R}\text{.}\)

*Constraints:*

To work out the relations among accelerations, we notice that tape unwinds smotthly on the pulley and does not slide over it. That would mean, the accelerations of the disks will come from both unwindind of their tapes and the spinning of the pulley.

Point \(y\) axis down and denote \(y\) acceleration components of the disks by \(a_1\) and \(a_2\text{,}\) and angular acceleration magnitudes by \(\alpha_1\) and \(\alpha_2\) of the disks and \(\alpha\) of the pulley. Looking at two instants, in the motion as shown in Figure 9.17.54. The rotation of the pulley adds AA'=BB' to disk 1 and subtracts CC'=BB'. In addition unwinding of tape adds A'A" to disk 1 and C'C" to disk2. Therefore, the accelerations are related as follows.

*Equations of Motion:*

The translational motion of the centers of the two disks give the following equations of motion.

The rotational motion about axis through the centers of pulley and the disks give us

*Solution of Equations of Motion:*

We solve Eqs. (9.17.3)-(9.17.9) simultaneously for seven unknowns, \(T_1,\ T_2,\ a_1,\ a_2,\ \alpha_1,\ \alpha_2,\ \alpha\text{.}\) I will illustrate elimintation steps to \(\alpha\) only. You can fill in the work for the others.

First elimintate \(a_1\) and \(a_2\) from Eqs. (9.17.5) and (9.17.6) using the constraint equations Eqs. (9.17.3) and (9.17.4).

Now, using Eqs. (9.17.7) and (9.17.8), we eliminate \(T_1\) and \(T_2\) from Eqs. (9.17.10), (9.17.11), and (9.17.9). We end up with three equations in \(\alpha_1\text{,}\) \(\alpha_2\text{,}\) and \(\alpha\text{.}\)

Using first of these for \(m_1R_1\alpha_1\) and second for \(m_2R_2\alpha_2\) in the third equation leads to \(\alpha\) of the pulley.