Question

Suppose the surface in the example is sloped as shown in the figure, but everything else remains the same. Determine the equation for the acceleration in this case using angular momentum and torque. Let a positive acceleration represent acceleration of m₂ up the surface. (Use the following as necessary: m₁, m₂, M, θ, and g.)

A sphere of mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley as shown in the figure. The radius of the pulley is R, and the mass of the thin rim is M. The spokes of the pulley have negligible m The block slides on a ass. frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects, using the concepts of angular and torque. SOLVE IT Conceptualize When the system is released, the block slides to hen the system is released the left, the sphere drops downward, and the pulley rotates the sphere moves downward and counterclockwise. This situation ar to proble ms we have he block moves to the left. solved earlier except that now we want to use an angular approach Categorize We identify the block, pulley, and sphere as a nonisolated system for angular momentum, subject to the external torque due o the gravitational force on the sphere. We shal calculate the angula m about axis that coincides with the axle of the pulley. The angular momentum of the syste udes that of two objects moving translationally (the sphere and the block) an d one object undergoing pure rotation (the pulley) Analyze At any instant of time, the sphere and the block have a ommon speed v, so the angular of the sphere is m1vR and that of the block is m2vR. 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 et's address the total externa orque acting on the system about he pulley axle. Because it has a moment arm of zero, the force exerted by the axle on the pulley does not contribute to th orque Further rmore, the normal force acting on the bloc s balanced by the gravitational force m2g, so these forces do not contribute to the orque. The gravitational force m19 act g on the sphere produces a torque about the axle equal agnitude o m19R, where R is the moment arm of the force about th Text m19R axle. This result is the total extennal torque about the pulley axle that is Write an expression for the total angula (1) L m1VR m2 (m1 m2 M)VR of the system dL Substitute this expression and the total Text external torque into the equation, the mathematical representation of the m19R m1 m2 solated system model for angula m19R m2 MDR dt Recognizing that dv/dt a, solve Equation (2) for a se the following as necessa m, M m1, m2, M, and g. Finalize When we evaluated the net torgue about the axle we did not in ude the forces thi he cord exerts on the objects because these forces are internal o the system under consideration ead, we analyzed the system as a whole. Only external torques contribute o the change in he system's angula MASTER IT Suppose the surface in the example is sloped as shown n the figure, b rything ns the e. Determine the equation for the acceleration in this case and torque. Let sing angular momentul accele ation of m2 up th e surface. Use he following as necessary: m1 posi ve acceleration represen m2, M, 6 and g misin(8) m. sin(6 M-m.

Answer 1

Consider that the velocity of $m_{1}$ is downwards and velocity of $m_{2}$ is up the inclined plane.

Angular momentum of system ($m_{1},m_{2},M$) is $L=m_{1}VR+m_{2}VR+MVR$

$\sum{\tau_{ext}}=\frac{dL}{dt}$

External forces are

on $m_{1}$ gravitational force $m_{1}g$ vertically downwards

on $m_{2}$ gravitational force $m_{2}g$ vertically downwards and Normal reaction $N$ perpendicular to the plane.

$N=m_{2}g\sin{\theta}$

component of gravitational force down the plane is $m_{2}g\cos{\theta}$

on $M$ gravitational force $Mg$ and support forces due to table.

Net torque is $\sum{\tau_{ext}}=m_{1}gR-m_{2}g\cos{\theta}R$

$\sum{\tau_{ext}}=m_{1}gR-m_{2}g\cos{\theta}R=\frac{dL}{dt}=\frac{d}{dt}(m_{1}VR+m_{2}VR+MVR)$

$m_{1}gR-m_{2}g\cos{\theta}R=(m_{1}R+m_{2}R+MR)a$ where $a=\frac{dV}{dt}$

$a=\frac{m_{1}g-m_{2}g\cos{\theta}}{m_{1}+m_{2}+M}$