A disc of radius 2; m and mass 100; kg rolls | vedclass

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(A) The work required to stop the disc is equal to the change in its total kinetic energy.Total kinetic energy of a rolling disc is given by $KE = KE_

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$3\; J$

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(A) The work required to stop the disc is equal to the change in its total kinetic energy.Total kinetic energy of a rolling disc is given by $KE = KE_{translational} + KE_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a disc,the moment of inertia $I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.Substituting these,$KE = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.Given: $m = 100\; kg$,$v = 20\; cm/s = 0.2\; m/s$.$KE = \frac{3}{4} 	imes 100 	imes (0.2)^2 = 75 	imes 0.04 = 3\; J$.Since the work done to stop the object is $W = \Delta KE = 0 - KE_{initial} = -3\; J$,the magnitude of work needed is $3\; J$.

$A$ disc of radius $2\; m$ and mass $100\; kg$ rolls on a horizontal floor. Its centre of mass has a speed of $20\; cm/s$. How much work is needed to stop it?

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