Question 11.4: A System of Objects A sphere of mass m1 and a block of mass ...
A System of Objects
A sphere of mass m_1 and a block of mass m_2 are connected by a light cord that passes over a pulley as shown in Figure 11.6. The radius of the pulley is R, and the mass of the thin rim is M. The spokes of the pulley have negligible mass. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects, using the concepts of angular momentum and torque.
Conceptualize When the system is released, the block slides to the left, the sphere drops downward, and the pulley rotates counterclockwise. This situation is similar to problems we have solved earlier except that now we want to use an angular momentum approach.
Categorize We identify the block, pulley, and sphere as a nonisolated system for angular momentum, subject to the external torque due to the gravitational force on the sphere. We shall calculate the angular momentum about an axis that coincides with the axle of the pulley. The angular momentum of the system includes that of two objects moving translationally (the sphere and the block) and one object undergoing pure rotation (the pulley).
Analyze At any instant of time, the sphere and the block have a common speed v, so the angular momentum of the sphere about the pulley axle is m_1 v R and that of the block is m_2 v R. At the same instant, all points on the rim of the pulley also move with speed v, so the angular momentum of the pulley is MvR.
Now let’s address the total external torque acting on the system about the pulley axle. Because it has a moment arm of zero, the force exerted by the axle on the pulley does not contribute to the torque. Furthermore, the normal force acting on the block is balanced by the gravitational force m_2 \overrightarrow{g}, so these forces do not contribute to the torque. The gravitational force m_1 \overrightarrow{g} acting on the sphere produces a torque about the axle equal in magnitude to m_1 g R, where R is the moment arm of the force about the axle. This result is the total external torque about the pulley axle; that is, \sum \tau_{\text {ext }}=m_1 g R.
Write an expression for the total angular momentum of the system:
(1) L=m_1 v R+m_2 v R+M v R=\left(m_1+m_2+M\right) v R
Substitute this expression and the total external torque into Equation 11.15, the mathematical representation of the nonisolated system model for angular momentum:
\sum \overrightarrow{\tau}_{\text{ext}}=\frac{d \overrightarrow{L}_{\text{tot}}}{d t} (11.15)
\begin{aligned} \sum \tau_{\text {ext }}&=\frac{d L}{d t}\\m_1 g R&=\frac{d}{d t}\left[\left(m_1+m_2+M\right) v R\right]\\ (2) m_1 g R&=\left(m_1+m_2+M\right) R \frac{d v}{d t}\end{aligned}Recognizing that dv/dt = a, solve Equation (2) for a:
(3) a=\frac{m_1 g}{m_1+m_2+M}
Finalize When we evaluated the net torque about the axle, we did not include the forces that the cord exerts on the objects because these forces are internal to the system under consideration. Instead, we analyzed the system as a whole. Only external torques contribute to the change in the system’s angular momentum. Let M → 0 in Equation (3) and call the result Equation A. Now go back to Equation (5) in Example 5.10, let θ → 0, and call the result Equation B. Do Equations A and B match? Looking at Figures 5.16 and 11.6 in these limits, should the two equations match?
(5) a=\left(\frac{m_2 \sin \theta-m_1}{m_1+m_2}\right) g
