IIT-JEE 2008 · Paper 2 · Physics Q41

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Passage

Paragraph for Question Nos. 39 to 41

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 (CM) at a distance $L$ from the wall. The disk rolls without slipping with velocity $\vec V_0=V_0\hat i$ . The coefficient of friction is $\mu$ .

A disk of radius R standing on a horizontal floor; its massless axle extends a distance d on either side of the centre, and from each end a spring runs back to a plate on the vertical wall (the spring attachments on the wall are 2d apart). The disk moves away from the wall with velocity V₀ along x.

The maximum value of $V_0$ for which the disk will roll without slipping is

\mu g\sqrt{\dfrac Mk}

\mu g\sqrt{\dfrac M{2k}}

\mu g\sqrt{\dfrac{3M}k}

\mu g\sqrt{\dfrac{5M}{2k}}

Answer and worked solution

Official answer: C

Friction needed. The springs supply $-2kx$ but the net force is $-\frac43kx$ , so friction is $f=+\frac23kx$ (opposing the springs).

Largest value. At maximum displacement $x_0=\frac{V_0}\omega$ :
$\frac23k\frac{V_0}\omega\le\mu Mg\implies V_0\le\frac{3\mu Mg}{2k}\sqrt{\frac{4k}{3M}}=\mu g\sqrt{\frac{3M}k}.$

(C)

Topics:Friction Rolling motion

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IIT-JEE 2008 · Paper 2, Physics question 41