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...

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 $\displaystyle{m}.$

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 $\displaystyle{m}.$

Denote the speed of a mass $\displaystyle{m}$ as $\displaystyle{V}.$ 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 $\displaystyle\frac{{{m}{{V}}^{{2}}}}{{2}}$ and the kinetic energy of a rotating pulley is $\displaystyle\frac{{{M}{{V}}^{{2}}}}{{4}}.$ 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:

$\displaystyle{m}{g{{h}}}=\frac{{{m}{{\left({h}'\right)}}^{{2}}}}{{2}}+\frac{{{M}{{\left({h}'\right)}}^{{2}}}}{{4}},$ or

$\displaystyle\frac{{{h}'}}{{\sqrt{{{h}}}}}=\sqrt{{\frac{{{m}{g}}}{{\frac{{m}}{{2}}+\frac{{M}}{{4}}}}}}=\sqrt{{{a}}}.$

Integrating this we obtain $\displaystyle{2}\sqrt{{{h}}}=\sqrt{{{a}}}{t}+{C},$ and C is obviously zero.

The solution is $\displaystyle{h}{\left({t}\right)}=\frac{{a}}{{4}}{{t}}^{{2}}$ (downwards), and so the acceleration is $\displaystyle\frac{{a}}{{4}}=\frac{{{m}{g}}}{{{2}{m}+{M}}}.$ This is the answer.

Notes

Thursday, December 19, 2013