Question

Solve the previous problem if the pulley has a moment of inertial I about its axis and the string does not slip over it.

Answer 1

Let us slove the problem by 'energy method', initial extension of the spring in the mean position $\delta =\frac{mg}{k.}$

During oscilation at any position 'x' below the equilibrium position let the velocity of 'm', be v and angular velocity of the pulley be ${}^{\prime }{\omega }^{\prime }.$ If r is the radius of the pulley, then

$v=r\omega$

At any instant, total energy = constant (For SHM)

$\therefore \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}K\left[\right(x+\delta {)}^2-{\delta }^2]-mgx=$ constant

$\Rightarrow \left(\frac{1}{2}\right)m{v}^2+\left(\frac{1}{2}\right)I{\omega }^2+\left(\frac{1}{2}\right)k{x}^2+kxd-mgx=$ constant

$\Rightarrow (\frac{1}{2}m{v}^2+\left(\frac{1}{2}\right)l(v^2+r^2)+\left(\frac{1}{2}\right))k{x}^2=$ constant $\delta =mg/k)$

Taking derivative of both sides with respect of 't',

$mv.\frac{dv}{dt}+\frac{1}{r^2}v\frac{dv}{dt}+kx\frac{dx}{dt}=0$

$\Rightarrow a\left(\frac{m+1}{r^2}\right)=-kx$

$(\therefore v=\frac{dx}{dt}\text{and}a=\frac{dv}{dt})$

$\Rightarrow \frac{a}{r}=\frac{k}{\frac{m+1}{r^2}}={\omega }^2$

$\Rightarrow T=2\pi \sqrt{\frac{m+1}{\frac{r^2}{k}}}$