A solid cylinder of mass 3 kg is rolling on | vedclass

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(D) At maximum compression,the translational and rotational motion of the solid cylinder stops momentarily.By the law of conservation of energy,the to

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$0.6$

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(D) At maximum compression,the translational and rotational motion of the solid cylinder stops momentarily.By the law of conservation of energy,the total kinetic energy of the rolling cylinder is converted into the potential energy of the spring.Total $K.E. = K.E._{trans} + K.E._{rot} = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$.For a solid cylinder,$I = \frac{1}{2} mR^2$ and for rolling without slipping,$\omega = \frac{v}{R}$.Substituting these,$K.E. = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{1}{2} mR^2) (\frac{v}{R})^2 = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2$.Equating this to the potential energy of the spring,$\frac{1}{2} kx^2 = \frac{3}{4} mv^2$.Given $m = 3 \ kg$,$v = 4 \ m/s$,and $k = 200 \ N/m$.$\frac{1}{2} 	imes 200 	imes x^2 = \frac{3}{4} 	imes 3 	imes (4)^2$.$100 x^2 = \frac{9}{4} 	imes 16 = 36$.$x^2 = \frac{36}{100} = 0.36$.$x = 0.6 \ m$.

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