IIT-JEE 2008 · Paper 2 · Physics Q39

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PhysicsSingle correct+4 / -1Question 39 of 66

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 net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is

-kx

-2kx

-\dfrac{2kx}3

-\dfrac{4kx}3

Answer and worked solution

Official answer: D

Energy of the rolling disk. Two springs give potential energy $2\cdot\frac12kx^2=kx^2$ . Rolling without slipping, the kinetic energy is $\frac12Mv^2+\frac12\cdot\frac12MR^2\cdot\frac{v^2}{R^2}=\frac34Mv^2$ .

Equation of motion. $\frac{d}{dt}\left(kx^2+\frac34Mv^2\right)=0$ gives $\frac32Ma=-2kx$ , so the net external force (springs plus friction) is
$Ma=-\frac43kx.$

(D)

Topics:Rolling motion Spring-mass systems

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IIT-JEE 2008 · Paper 2, Physics question 39