# The pulley shown in the figure has a moment of inertia I about its axis and mass m. Find the time period of vertical oscillation of its center of mass. The spring has spring constant K and the string does not slip on the pulley?

Mar 20, 2018

See below.

#### Explanation:

Assuming the pulley of uniform material.

$I = \frac{1}{2} m {r}^{2}$

Pulley dynamics

$m \alpha = m g + {T}_{1} + {T}_{2}$
$I \dot{\omega} = r \left({T}_{1} - {T}_{2}\right)$

then solving

$\left\{\begin{matrix}{T}_{1} + {T}_{2} = - m g + m \alpha \\ {T}_{1} - {T}_{2} = \frac{I}{r} \dot{\omega}\end{matrix}\right.$

we obtain

${T}_{1} = \frac{1}{2} \left(- m g + m \alpha + \frac{I}{r} \dot{\omega}\right)$

but

${T}_{1} = - k x , \alpha = \frac{\ddot{x}}{2}$ and $\dot{\omega} = \frac{\ddot{x}}{r}$ so

$\left(\frac{m}{2} + \frac{m}{2}\right) \ddot{x} + 2 k x - m g = 0$ or

$m \ddot{x} + 2 k x - m g = 0$

now considering the homogeneous differential equation

$\frac{m}{2 k} {\ddot{x}}_{h} + {x}_{h} = 0$

obtaining

$\Omega = \sqrt{\frac{2 k}{m}}$ and

period $\boldsymbol{T} = \frac{2 \pi}{\Omega} = 2 \pi \sqrt{\frac{m}{2 k}}$