Question

A block of mass $m$ hangs by means of a string which goes over a pulley of mass $M$ and moment of inertia $I$, as shown in the diagram. The string does not slip relative to the pulley. Find the frequency of small oscillations.

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Accepted Answer

At the time of equilibrium: $K x_{0} = 2 m g + M g \dots (i)$

If the block is depressed by $x,$ the pulley owing to the constraint is depressed by $x / 2 .$ If the acceleration of the block is $a,$ then the acceleration of the pulley should be $a / 2 .$

Writing equation of motion

For block: $T - m g = m a \dots (ii)$

For pulley: $K (x_{0} + \frac{x}{2}) - T - T^{'} - M g = \frac{M a}{2} \dots (iii)$

The acceleration of point $' P'$ should be zero as it is directly connected to a fixed point.

Hence, $α R = \frac{a}{2} \Rightarrow α = \frac{a}{2 R} \dots (iv)$

Now writing torque equation for pulley: $T^{'} R - T R = I α = I \cdot \frac{a}{2 R}$

or $T^{'} - T = \frac{I a}{2 R^{2}} \dots (v)$

Using above equations, we get $a = \frac{K x}{[4 m + M + \frac{I}{R^{2}}]}$

As the acceleration vector and displacement vectors both are opposite to each other, hence above equation can be written in vector form as, $\vec{a} = \frac{K}{[4 m + M + \frac{I}{R^{2}}]} \cdot \vec{x}$

Comparing with equation: $\vec{a} = - ω^{2} \vec{x};$ we get

$ω = \frac{K}{(4 m + M + \frac{I}{R^{2}})}$ or $ω = \sqrt{\frac{K}{(4 m + M + \frac{I}{R^{2}})}}$

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

Method $2 :$ Point $' P'$ is the instantaneous center of rotation. Let the pulley be rotated by a small angle $θ$ about an axis passing through $P .$

Due to this rotation, the spring will be elongated by a length $R θ .$ This increased elongation of spring will provide a restoring torque about point $P .$ Now writing torque equation about $P .$

$(Δ F_{S}) R = I_{P} α$

$(K R θ) \cdot R = (I + M R^{2} + m (2 R)^{2}) α$

or $α = \frac{K R^{2} θ}{(I + M R^{2} + 4 M R^{2})} \dots (i)$

Here we are not writing the torques due to weight of the pulley and block, as these torques are not responsible for providing restoring torques.

Equation (i) can be written in vector form, $\vec{α} = \frac{K}{(4 m + M + \frac{I}{R^{2}})} \vec{θ}$

Comparing with standard equation of angular SHM, $\vec{α} = - ω^{2} \vec{θ}$

We get $ω^{2} = \frac{K}{(4 m + M + \frac{I}{R^{2}})}$ or $T = 2 π \sqrt{\frac{(4 m + M + \frac{I}{R^{2}})}{K}}$

The time period of oscillation of the pulley and block will be same.

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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