Question

The pulley shown in figure has a moment of inertia I about its axis and mass m. Find the time period of vertical oscillation of its centre of mass. The spring has spring constant k and the string does not slip over the pulley.

• A. $2\pi \sqrt{\frac{(\frac{I}{r^2}+m)}{4k}}$
• B. $2\pi \sqrt{\frac{4k}{(\frac{I}{r^2}+m)}}$
• C. $\sqrt{\frac{(\frac{I}{r^2}+m)}{4k}}$
• D. none of these

Answer 1

The correct option is A

$2\pi \sqrt{\frac{(\frac{I}{r^2}+m)}{4k}}$

In the position of equilibrium we have 2T = mg

Where T is the tension in each string

$k{y}_0=T\Rightarrow k{y}_0=\frac{mg}{2}:y_0=\frac{mg}{2k}$

Also if the extension of the spring is y then

If the CM of the pulley goes down by x the string must extend by an amount of 2x (x on both the sides) but string is of constant length, the spring must extend by 2x. At this random position the pulley is also rotating (has on angular velocity $\omega$)

$v=\frac{1}{2}l{\omega }^2+\frac{1}{2}m{v}^2-mgx+\frac{1}{2}k{(\frac{mg}{2k}+2x)}^2$

So in this position the total energy

Now we also know that energy is conserved.

So $\frac{du}{dt}=0$

$\Rightarrow \frac{d}{dt}[\frac{1}{2}I{\left(\frac{v}{r}\right)}^2+\frac{m}{2}{v}^2-mgx+\frac{1}{2}k{(\frac{mg}{2k}+2x)}^2]$

$\Rightarrow (\frac{I}{r^2}+m)v\frac{dv}{dt}+4kxv=0$

$\Rightarrow \frac{dv}{dt}=\frac{-4kx}{(\frac{I}{r^2}+m)}$

$a=-{\omega }^{2}x$ $\Rightarrow {\omega }^2=\frac{4k}{(\frac{I}{r^2}+m)}$

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{(\frac{I}{r^2}+m)}{4k}}$