A solid cylinder of mass 2 ; kg and radius 50 | vedclass

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(D) The total kinetic energy of a rolling body is $K = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.For a solid cylinder,the radius of gyration $K$ satisfies

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$2.4$

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(D) The total kinetic energy of a rolling body is $K = \frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$.For a solid cylinder,the radius of gyration $K$ satisfies $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.By the law of conservation of energy,the initial kinetic energy equals the final potential energy at the maximum height $h$:$\frac{1}{2}mv^2(1 + \frac{1}{2}) = mgh$$\frac{1}{2}v^2(\frac{3}{2}) = gh$Given $v = 4 \; m/s$ and taking $g = 10 \; m/s^2$:$\frac{1}{2} 	imes 16 	imes \frac{3}{2} = 10h$$12 = 10h \Rightarrow h = 1.2 \; m$.The distance travelled along the incline $\ell$ is related to height $h$ by $h = \ell \sin 30^{\circ}$.$\ell = \frac{h}{\sin 30^{\circ}} = \frac{1.2}{0.5} = 2.4 \; m$.

$A$ solid cylinder of mass $2 \; kg$ and radius $50 \; cm$ rolls up an inclined plane of angle of inclination $30^{\circ}$. The centre of mass of the cylinder has a speed of $4 \; m/s$. The distance travelled by the cylinder on the inclined surface will be ...... $m$.

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