# Q7.27     Prove the result that the velocity $v$ of translation of a rolling body (like a ring, disc,                cylinder or sphere) at the bottom of an inclined plane of a height $h$  is given by               $v^{2}=\frac{2gh}{(1+k^{2}/R^{2)}}$              using dynamical consideration (i.e. by consideration of forces and torques). Note $k$ is              the radius of gyration of the body about its symmetry axis, and $R$  is the radius of the              body. The body starts from rest at the top of the plane.

Consider the given situation :

The total energy when the object is at the top (potential energy)  =  mgh.

Energy when the object is at the bottom of the plane :

$E_b\ =\ \frac{1}{2}I\omega^2\ +\ \frac{1}{2}mv^2$

Put    $I\ =\ mk^2$       and      $v\ =\ \omega r$, we get :

$E_b\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )$

By the law of conservation of energy we can write :

$mgh\ =\ \frac{1}{2}mv^2\left ( 1\ +\ \frac{k^2}{r^2} \right )$

or

$v\ =\ \frac{2gh}{\left ( 1\ +\ \frac{k^2}{r^2} \right )}$                                        .

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