Question

The pulley shown in figure has a radius $r,$ moment of inertia $I$ about its axis and mass $m .$ Find the time period of vertica oscillations of its center of mass. The spring constant of spring is $K$ and the spring does not slip over the pulley.

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Accepted Answer

Here point $P$ is the instantaneous center of rotation. Let the pulley is rotated in clockwise sense by a small angle $θ$ beyond its equilibrium position. Because of this rotation the spring will be further elongated by a length $2 R θ .$ This increased length will develop an additional force $K (2 R θ),$ which will provide a restoring torque about an axis passing through $P .$ Now writing torque equation about $P .$

Now writing torque equation about $P, τ_{P} = I_{P} α$

$I_{P} α = [K (2 R θ)] \times 2 R \Rightarrow [I + M R^{2}] α = 4 K R^{2} θ$

Hence, $α = \frac{4 K}{[M + \frac{I}{R^{2}}]} θ$

As angular acceleration $(α)$ and angular displacement $(θ),$ both are directed in opposite direction. Hence we can write the equation, $α = - \frac{4 K}{(M + \frac{I}{R^{2}})} θ$ $\Rightarrow ω^{2} = \frac{4 K}{(M + \frac{I}{R^{2}})}$

Time period $T = 2 π \sqrt{\frac{(M + \frac{I}{R^{2}})}{4 K}} \Rightarrow T = π \sqrt{\frac{(M + \frac{I}{R^{2}})}{K}}$

B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS

Questions from B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS with Hints & Solutions

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