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Section 9.17 Rotation Bootcamp

Subsection 9.17.1 Describing Motion

Subsection 9.17.2 Rotational Speed and Velocity

Subsection 9.17.3 Rotational Acceleration

Subsection 9.17.4 Constant Rotational Acceleration

Subsection 9.17.5 Angular Momentum

Subsection 9.17.6 Moment of Inertia

Subsection 9.17.7 Torque

Subsection 9.17.8 Rotational Second Law

Subsection 9.17.9 Rotational Impulse

Subsection 9.17.10 Rotational Impulse and Angular Momentum

Subsection 9.17.11 Conservation of Angular Momentum

Subsection 9.17.12 Collision and Angular Momentum

Subsection 9.17.13 Rotational Work

Subsection 9.17.14 Rotational Kinetic Energy

Subsection 9.17.15 Rolling Motion

Subsection 9.17.16 Miscellaneous

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.

Figure 9.17.44.

(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?

Hint 1 (a)

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

Hint 2 (b)

(b) The torques are balanced.

Answer

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

Solution 1 (a)

(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.

\begin{equation*} I\alpha = \tau = T\, \dfrac{2}{3} R, \end{equation*}

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

\begin{equation*} I = \dfrac{1}{2}MR^2. \end{equation*}

Therefore magnitude of the angular acceleration is

\begin{equation*} \alpha = \dfrac{4}{3}\, \dfrac{T}{MR}. \end{equation*}

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

Solution 2 (b)

(b) Now, the two torques are balanced.

\begin{equation*} T_0\, \dfrac{2}{3} R = F_k\, r. \end{equation*}

Therefore,

\begin{equation*} F_k = \dfrac{2}{3}\, \dfrac{R}{r}\, T_0. \end{equation*}

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.

Figure 9.17.46.

(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.

Hint

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

Answer

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

Solution 1 (a)

(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

\begin{equation*} I\alpha = \dfrac{M g L}{2},\ \ \ I = \dfrac{1}{3}\,ML^2. \end{equation*}

Therefore,

\begin{equation*} \alpha = \dfrac{3}{2}\, \dfrac{g}{L}. \end{equation*}

Figure 9.17.47.
Solution 2 (b)

(b) Here, the lever arm changes to

\begin{equation*} r_{\perp} = \dfrac{L}{2}\,\sin\,\theta, \end{equation*}

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

\begin{equation*} I\alpha = \dfrac{M g L}{2}\,\sin\,\theta,\ \ \ I = \dfrac{1}{3}\,ML^2. \end{equation*}

Therefore,

\begin{equation*} \alpha = \dfrac{3}{2}\, \dfrac{g}{L}\, \sin\,\theta. \end{equation*}

Figure 9.17.48.

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:

Figure 9.17.50.

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 \(1.01\text{ m/s}\text{.}\)

Hint

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

Answer

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

Solution

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.

Figure 9.17.51. Figure for Problem 9.17.49 solution.

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

\begin{equation*} y = \frac{l}{2} \cos\theta. \end{equation*}

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.

\begin{equation} \frac{dy}{dt} = -\frac{l}{2} \sin\theta\frac{d\theta}{dt}. \label{eq-ch9-pr-9-2}\tag{9.17.1} \end{equation}

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,

\begin{equation*} \frac{1}{2} m \left( \frac{dy}{dt}\right)^2 + \frac{1}{2} I_0 \left( \frac{d\theta}{dt}\right)^2 + m g y = m g \frac{l}{2}, \end{equation*}

where

\begin{equation*} I_0 = \frac{1}{12} m l^2. \end{equation*}

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

\begin{equation} \left( \frac{dy}{dt}\right)^2 = \frac{3g l (1-\cos\theta)}{3+ \textrm{cosec}^2\theta}.\label{eq-ch9-pr-9-4}\tag{9.17.2} \end{equation}

Therefore, the velocity of the CM is

\begin{equation*} \vec v = \frac{dy}{dt} \hat j = - \left[\frac{3g l (1-\cos\theta)}{3+ \textrm{cosec}^2\theta} \right]^{1/2}\ \hat j, \end{equation*}

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

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.

Figure 9.17.53. Figure for Problem 9.17.52.
Hint

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

Answer

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

Solution

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.

\begin{align} a_1 \amp = R_1\alpha_1 + R \alpha, \label{one-pulley-two-disk-cons-a1}\tag{9.17.3}\\ a_2 \amp = R_2\alpha_2 - R \alpha. \label{one-pulley-two-disk-cons-a2}\tag{9.17.4} \end{align}
Figure 9.17.54. Figure for Problem 9.17.52 constraints.

Equations of Motion:

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

\begin{align} m_1a_1 \amp = m_1 g - T_1, \label{one-pulley-two-disk-eom-d1}\tag{9.17.5}\\ m_2a_2 \amp = m_2 g - T_2. \label{one-pulley-two-disk-eom-d2}\tag{9.17.6} \end{align}

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

\begin{align} \amp T_1 R_1 = \dfrac{1}{2} m_1 R_1^2 \alpha_1, \label{one-pulley-two-disk-eom-rot-d1}\tag{9.17.7}\\ \amp T_2 R_2 = \dfrac{1}{2} m_2 R_2^2 \alpha_2, \label{one-pulley-two-disk-eom-rot-d2}\tag{9.17.8}\\ \amp T_1 R - T_2 R = \dfrac{1}{2} M R^2 \alpha. \label{one-pulley-two-disk-eom-rot-p}\tag{9.17.9} \end{align}

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).

\begin{align} \amp m_1R_1\alpha_1 + m_1 R \alpha = m_1 g - T_1, \label{one-pulley-two-disk-eom-d1-a1-elim}\tag{9.17.10}\\ \amp m_2R_2\alpha_2 - m_2 R \alpha = m_2 g - T_2. \label{one-pulley-two-disk-eom-d2-a2-elim}\tag{9.17.11} \end{align}

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 \(\alpha\text{.}\)

\begin{align} \amp \frac{3}{2} m_1R_1\alpha_1 + m_1 R \alpha = m_1 g, \label{one-pulley-two-disk-eom-d1-a1-T1-elim}\tag{9.17.12}\\ \amp \frac{3}{2} m_2R_2\alpha_2 - m_2 R \alpha = m_2 g. \label{one-pulley-two-disk-eom-d2-a2-T2-elim}\tag{9.17.13}\\ \amp m_1R_1\alpha_1 -m_2R_2\alpha_2 - M R \alpha = 0. \label{one-pulley-two-disk-eom-rot-p-T1-T2-elim}\tag{9.17.14} \end{align}

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.

\begin{equation*} \alpha = \left( \dfrac{m_1 - m_2}{ m_1 + m_2 + \frac{3}{2}M } \right)\ \dfrac{g}{R}. \end{equation*}