1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

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

Open in App
Solution

## Let us try to solve the problem using energy method. If δ is the displacement from the mean position then, the initial extension of the spring from the mean position is given by, $\delta =$$\frac{mg}{k}$ Let x be any position below the equilibrium during oscillation. Let v be the velocity of mass m and ω be the angular velocity of the pulley. If r is the radius of the pulley then v = rω As total energy remains constant for simple harmonic motion, we can write: $\frac{1}{2}M{v}^{2}+\frac{1}{2}I{\omega }^{2}+\frac{1}{2}k\left[\left(x+\delta {\right)}^{2}-{\delta }^{2}\right]-Mgx=\mathrm{Constant}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}M{v}^{2}+\frac{1}{2}I{\omega }^{2}+\frac{1}{2}k{x}^{\mathit{2}}+kxd\mathit{-}Mgx=\mathrm{Constant}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}M{v}^{2}+\frac{1}{2}I\left(\frac{{v}^{2}}{{r}^{2}}\right)+\frac{1}{2}k{x}^{\mathit{2}}=\mathrm{Constant}\left[\because \delta =\frac{Mg}{k}\right]\phantom{\rule{0ex}{0ex}}$ By taking derivatives with respect to t, on both sides, we have: $Mv.\frac{\mathrm{d}v}{\mathrm{d}t}+\frac{I}{{r}^{2}}v.\frac{\mathrm{d}v}{\mathrm{d}t}+kx\frac{\mathrm{d}x}{\mathrm{d}t}=0\phantom{\rule{0ex}{0ex}}Mva+\frac{I}{{r}^{2}}va+kxv=0\left(\because v=\frac{\mathrm{d}x}{\mathrm{d}t}\mathrm{and}a=\frac{\mathrm{d}v}{\mathrm{d}t}\right)\phantom{\rule{0ex}{0ex}}a\left(M+\frac{I}{{r}^{2}}\right)=-kx\phantom{\rule{0ex}{0ex}}⇒\frac{a}{x}=\frac{k}{M+\frac{I}{{r}^{2}}}={\omega }^{2}\phantom{\rule{0ex}{0ex}}T=\frac{2\pi }{\omega }\phantom{\rule{0ex}{0ex}}⇒T=2\pi \sqrt{\frac{M+\frac{I}{{r}^{2}}}{k}}$

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Centre of Mass in Galileo's World
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program