Question

A uniform spherical shell of mass $M$ and radius $R$ rotates about a vertical axis on the frictionless bearing. A massless cord passes around the equator of the shell, over a pulley of rotational inertia $I$ and radius $r$ and it attached to a small object of mass $m$ that is otherwise free to fall under the influence of gravity. There is no friction of pulley's axle; the cord does not slip on the pulley. What is the speed of the object after it has fallen a distance $h$ from rest? Use work - energy considerations.

Answer 1

$dV=2\pi R\sin \left(a\right)tRda$

$a\to$colatitude, $t\to$wall thickness

$\therefore r=R\sin \left(a\right)$

The density$\to \frac{M}{4\pi R^{2}t}$

Inertia of the sphere

${\int }_0^{\pi }{MR}^2{\sin }^{3}ada=\frac{2}{3}{MR}^2$

Angular velocity of sphere$:-w_s=\frac{V}{R}$

Angular velocity of pully$:-w_p=\frac{v}{r}$

$\frac{\left(\frac{2}{3}{MR}^2\right)}{2}+\frac{I{\left(\frac{v}{r}\right)}^2}{2}+\frac{{mv}^2}{2}+mgh=0$

$(\frac{M}{3}+\frac{I}{2r^2}+\frac{m}{2})V^2+mgh=0$

Energy equation for a simple mass

$V^2+2Gh=0$

For this simple case,

$h\left(t\right)=\frac{G{t}^2}{2}$

$V\left(h\right)=\sqrt{2Gh}$

$\therefore 2G=\frac{mg}{(\frac{M}{3}+\frac{I}{2r^2}+\frac{m}{2})}$

$\therefore V\left(h\right)=\sqrt{\frac{mgh}{\frac{M}{3}+\frac{I}{2r^2}+\frac{m}{2}}}$