A uniform spherical shell of mass $M$ and radius $R$ rotates about a vertical axis on frictionless bearing. A massless cord passes around the equator of the shell, over a pulley of rotational inertia $I$ and radius $R$ and is 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.

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Answer 1

By conservation of energy, when object falls by a distance of h.
$\frac{1}{2} m v^{2} + \frac{1}{2} I_{s h e l l} ω_{s h e l l}^{2} + \frac{1}{2} I_{p u l l e y} = m g h$
$v = R ω$
$\frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{3} M R^{2}) \times (\frac{v}{R})^{2} + \frac{1}{2} I_{p} (\frac{v}{r _{p}})^{2} = m g h$
$v^{2} (m + \frac{2}{3} M + \frac{I _{p}}{r _{p}^{2}}) = 2 m g h$
$v = \frac{2 m g h}{( m + \frac{2}{3} M + \frac{I _{p}}{r _{p}^{2}} )}$

Answer 2

By conservation of energy, when object falls by a distance of h.

\frac{1}{2} m v^{2} + \frac{1}{2} I_{s h e l l} ω_{s h e l l}^{2} + \frac{1}{2} I_{p u l l e y} = m g h

v = R ω

\frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{3} M R^{2}) \times (\frac{v}{R})^{2} + \frac{1}{2} I_{p} (\frac{v}{r _{p}})^{2} = m g h

v^{2} (m + \frac{2}{3} M + \frac{I _{p}}{r _{p}^{2}}) = 2 m g h

v = \frac{2 m g h}{( m + \frac{2}{3} M + \frac{I _{p}}{r _{p}^{2}} )}