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

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Question

(A) For a solid cylinder rolling down an inclined plane,the total potential energy $Mgh$ is converted into translational kinetic energy and rotational

Vedclass · Accepted Answer

$\frac{Mgh}{3}$

Vedclass · Answer

(A) For a solid cylinder rolling down an inclined plane,the total potential energy $Mgh$ is converted into translational kinetic energy and rotational kinetic energy.Total energy $E = Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$.For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the rolling condition is $v = R\omega$.Substituting these into the energy equation:$Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$$Mgh = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2 = \frac{3}{4} Mv^2$.Thus,$Mv^2 = \frac{4}{3} Mgh$.The rotational kinetic energy $K_{rot} = \frac{1}{2} I\omega^2 = \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2 = \frac{1}{4} Mv^2$.Substituting $Mv^2 = \frac{4}{3} Mgh$ into the expression for $K_{rot}$:$K_{rot} = \frac{1}{4} (\frac{4}{3} Mgh) = \frac{Mgh}{3}$.

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

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