Thursday, December 19, 2013

A string is wrapped around a pulley (a solid disk) of mass M and radius R, and is connected to a mass m. Solve for acceleration of m using...

Hello!


I suppose that we ignore friction and the weight of a string, and that the system starts from rest. I'll use the downward y-axis starting from the initial position of `m.`


Denote the speed of a mass `m` as `V.` Then the outer edge of a pulley will have the same linear speed. It is known that the kinetic energy of a linearly moving mass m is `(m V^2)/2` and the kinetic energy of...

Hello!


I suppose that we ignore friction and the weight of a string, and that the system starts from rest. I'll use the downward y-axis starting from the initial position of `m.`


Denote the speed of a mass `m` as `V.` Then the outer edge of a pulley will have the same linear speed. It is known that the kinetic energy of a linearly moving mass m is `(m V^2)/2` and the kinetic energy of a rotating pulley is `(M V^2)/4.` The potential energy of a mass m is -mgh (it moves down), and of course V(t)=h'(t) (the speed is the derivative of the displacement).


So we obtain a simple differential equation:


`mgh=(m (h')^2)/2+(M (h')^2)/4,` or


`(h')/(sqrt(h))=sqrt((mg)/(m/2+M/4))=sqrt(a).`


Integrating this we obtain `2sqrt(h)=sqrt(a) t+C,` and C is obviously zero. 


The solution is `h(t)=a/4 t^2` (downwards), and so the acceleration is `a/4=(mg)/(2m+M).` This is the answer. 

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