The maximum value of for which the disk will roll without slipping is – [2008]

Question

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 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 $μ$ . [2008]

Q. The maximum value of $V_{0}$ for which the disk will roll without slipping is –

Smart Achievers · Accepted Answer

(3)

Mechanical energy is conserved in case of pure rolling motion

$∴$ $\frac{1}{2} M v_{0}^{2} + \frac{1}{2} (\frac{1}{2} M R^{2}) {(\frac{v_{0}}{R})}^{2} = 2 [\frac{1}{2} k x_{\max}^{2}]$

$∴$ $x_{\max} = \sqrt{\frac{3 M}{4 k}} v_{0}$

$F_{\max} = μ M g = \frac{2 k x_{\max}}{3} = \frac{2 k}{3} \sqrt{\frac{3 M}{4 k} v_{0}}$

$∴$ $v_{0} = μ g \sqrt{\frac{3 M}{k}}$

Solution

1 $μ g \sqrt{\frac{M}{k}}$

2 $μ g \sqrt{\frac{M}{2 k}}$

3 $μ g \sqrt{\frac{3 M}{k}}$

4 $μ g \sqrt{\frac{5 M}{2 k}}$

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Solution

1 μgMk

2 μgM2k

3 μg3Mk

4 μg5M2k

Ans. (3) Mechanical energy is conserved in case of pure rolling motion ∴ 12Mv02+12(12MR2)(v0R)2=2[12kxmax2] ∴ xmax=3M4kv0 Fmax=μMg=2kxmax3=2k33M4kv0 ∴ v0=μg3Mk

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1 $μ g \sqrt{\frac{M}{k}}$

2 $μ g \sqrt{\frac{M}{2 k}}$

3 $μ g \sqrt{\frac{3 M}{k}}$

4 $μ g \sqrt{\frac{5 M}{2 k}}$