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

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Question

(A) For a solid cylinder rolling down an inclined plane,the total energy at the top is potential energy $PE = Mgh$. At the bottom,the total energy is

Vedclass · Accepted Answer

$\frac{3 v^{2}}{4 g}$

Vedclass · Answer

(A) For a solid cylinder rolling down an inclined plane,the total energy at the top is potential energy $PE = Mgh$. At the bottom,the total energy is the sum of translational kinetic energy and rotational kinetic energy: $KE = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$. For a solid cylinder,the moment of inertia $I = \frac{1}{2} MR^2$ and the rolling condition is $v = R\omega$,so $\omega = \frac{v}{R}$. Substituting these into the energy equation: $Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} MR^2) (\frac{v}{R})^2$. $Mgh = \frac{1}{2} Mv^2 + \frac{1}{4} Mv^2$. $Mgh = \frac{3}{4} Mv^2$. Solving for height $h$: $h = \frac{3 v^2}{4 g}$.

$A$ solid cylinder of mass $M$ and radius $R$ rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity $v$. What is the height of the inclined plane? (Assume $g$ is the acceleration due to gravity.)

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