A circular disc is rolling on a horizontal pl | vedclass

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Question

(A) The total kinetic energy $(KE_{total})$ of a rolling body is the sum of its translational and rotational kinetic energies:$KE_{total} = KE_{trans}

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) The total kinetic energy $(KE_{total})$ of a rolling body is the sum of its translational and rotational kinetic energies:$KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a circular disc,the moment of inertia $I = \frac{1}{2}mR^2$ and for pure rolling,$\omega = \frac{v}{R}$.Thus,$KE_{rot} = \frac{1}{2} (\frac{1}{2}mR^2) (\frac{v}{R})^2 = \frac{1}{4}mv^2$.$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.We know $KE_{trans} = \frac{1}{2}mv^2$.Therefore,the ratio is $\frac{KE_{trans}}{KE_{total}} = \frac{\frac{1}{2}mv^2}{\frac{3}{4}mv^2} = \frac{2}{3}$.Given $KE_{total} = 300 \ J$,the translational kinetic energy is $KE_{trans} = \frac{2}{3} 	imes 300 \ J = 200 \ J$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

$A$ circular disc is rolling on a horizontal plane. Its total kinetic energy is $300 \ J$. What is its translational kinetic energy?

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