Paragraph: A uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless…

Paragraph: A uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(C M)$ at a distance Lfrom the wall. The disk rolls without slipping with velocity $\mathbf{v}_0=v_0 \hat{\mathbf{i}}$ The coefficient of friction is $\mu$.
Question: The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is

1. $-k x$
2. $-2 k x$
3. $-\frac{2 k x}{3}$
4. $-\frac{4 k x}{3}$

Solution

$\therefore \quad \frac{2 k x-f}{M}=R\left[\frac{f \cdot R}{\frac{1}{2} M R^2}\right]$ Solving this equation, we get $f=\frac{2 k x}{3}$ $ \therefore\left|F_{\text {net }}\right|=2 k x-f=2 k x-\frac{2 k x}{3}=\frac{4 k x}{3} $ This is opposite to displacement. $ \therefore \quad F_{\text {net }}=-\frac{4 k x}{3} $ $\therefore$ correct option is (d). `

Asked in: JEE Advanced 2008 (Paper 2)

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