A solid cylinder of mass M and radius R rolls | vedclass

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Question

(A) By the law of conservation of energy,the total potential energy at the top is equal to the sum of translational and rotational kinetic energy at t

Vedclass · Accepted Answer

$\frac{Mgh}{3}$

Vedclass · Answer

(A) By the law of conservation of energy,the total potential energy at the top is equal to the sum of translational and rotational kinetic energy at the bottom:$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the condition for pure rolling is $v = R\omega$.Substituting these into the energy equation:$Mgh = \frac{1}{2} M(R\omega)^2 + \frac{1}{2} (\frac{1}{2} MR^2)\omega^2$$Mgh = \frac{1}{2} MR^2\omega^2 + \frac{1}{4} MR^2\omega^2 = \frac{3}{4} MR^2\omega^2$Thus,the rotational kinetic energy $K_{rot} = \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{1}{2} MR^2)\omega^2 = \frac{1}{4} MR^2\omega^2$.From the energy equation,$MR^2\omega^2 = \frac{4}{3} Mgh$.Substituting this into the expression for $K_{rot}$:$K_{rot} = \frac{1}{4} (\frac{4}{3} Mgh) = \frac{Mgh}{3}$.

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. When it reaches the foot of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).

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