A uniform disc of mass 0.5,kg and radius r is | vedclass

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(C) Given: mass $m = 0.5\,kg$,initial velocity $u = 18\,m/s$,coefficient of friction $\mu = 0.3$,$g = 10\,m/s^2$,time $t = 2\,s$.The acceleration of t

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$54$

Vedclass · Answer

(C) Given: mass $m = 0.5\,kg$,initial velocity $u = 18\,m/s$,coefficient of friction $\mu = 0.3$,$g = 10\,m/s^2$,time $t = 2\,s$.The acceleration of the disc due to friction is $a = -\mu g = -0.3 	imes 10 = -3\,m/s^2$.The velocity of the center of mass at $t = 2\,s$ is $v = u + at = 18 - 3 	imes 2 = 12\,m/s$.For pure rolling,the condition is $v = \omega r$,so $\omega = v/r$.The total kinetic energy $KE$ is the sum of translational and rotational kinetic energy:$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since $I = \frac{1}{2}mr^2$ for a disc,we have:$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.Substituting the values:$KE = \frac{3}{4} 	imes 0.5 	imes (12)^2 = \frac{3}{4} 	imes 0.5 	imes 144 = 3 	imes 0.5 	imes 36 = 54\,J$.

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