Question 17.SP.10: A 10-lb uniform disk is attached to the shaft of a motor mou...
A 10-lb uniform disk is attached to the shaft of a motor mounted on arm AB that is free to rotate about the vertical axle CD. The arm-and-motor unit has a moment of inertia of 0.032 lb·ft·s² about axle CD. Knowing that the system is initially at rest, determine the angular velocities of the arm and of the disk when the motor reaches a speed of 360 rpm.
STRATEGY: Since you have two times––when the system starts from rest and when the motor has reached a speed of 360 rpm––use the conservation of angular momentum. You cannot use the conservation of energy because the motor converts electrical energy into mechanical energy.
MODELING: Choose the arm AB, the motor, and the disk to be your system and model them as rigid bodies. The impulse–momentum diagram for this system is shown in Fig. 1.
Moments of Inertia. The mass moment of inertia of the arm and motor about the axle is I_A=0.032 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2, and the mass moment of inertia of disk B about its center of mass is
\bar{I}_B=\frac{1}{2} \frac{W}{g} r^2=\frac{1}{2}\left(\frac{10}{32.2}\right)\left(\frac{5}{12}\right)^2=0.02696 \mathrm{lb} \cdot \mathrm{ft} \cdot \mathrm{s}^2
ANALYSIS:
Principle of Impulse and Momentum. Apply the principle of impulse and momentum for this system between time t_1 (when the system is at rest) and t_2 (when the motor has an angular velocity of 360 rpm)
\text { Syst Momenta }{ }_1 + \text { Syst Ext Imp } _{1 \rightarrow 2}=\text { Syst Momenta }{ }_2
Taking moments about A gives
+\circlearrowleft \text { moments about } A: \quad 0 + 0=\left(m_B v_B\right) l_{A B} + I_A \omega_{A B} + \bar{I}_B \omega_B (1)
Kinematics. You can relate the velocity of B to the angular velocity of AB using
v_B=l_{A B} \omega_{A B}=\frac{6}{12} \omega_{A B} (2)
The velocity of the motor is \omega_M=360 \mathrm{rpm}=12 \pi \mathrm{rad} / \mathrm{s}, which is the angular velocity of the disk relative to the arm. Thus,
\omega_B=\omega_{A B} + \omega_M (3)
Substituting Eqs. (2) and (3) into Eq. (1) and solving for v_{AB} gives
\begin{gathered}\left(m_B l_{A B}^2 + I_A\right) \omega_{A B} + \bar{I}_B\left(\omega_{A B} + \omega_M\right)=0 \\{\left[\left(\frac{10}{32.2}\right)\left(\frac{6}{12}\right)^2 + 0.032\right] \omega_{A B} + 0.02696\left(\omega_{A B} + 12 \pi\right)=0} \\\omega_{A B}=-7.44 \mathrm{rad} / \mathrm{s}\end{gathered}
\boldsymbol{\omega}_{A B}=71.0 \mathrm{rpm} \circlearrowright
The angular velocity of the disk is
\omega_B=-7.44 + 12 \pi=30.26 \mathrm{rad} / \mathrm{s}
\boldsymbol{\omega}_B=289 \mathrm{rpm}\circlearrowleft
REFLECT and THINK: When the motor spins the disk counterclockwise (as viewed from above), the arm AB rotates in a clockwise direction. One key to solving this problem is recognizing that the angular velocity of the motor is the relative angular velocity of the disk with respect to the bar.