Prove 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 height $h$ is given by $v^{2} = \frac{2gh}{1 + k^{2}/R^{2}}$ using dynamical considerations (i.e.,by considering 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.

(N/A) Let a body of mass $m$,radius $R$,and radius of gyration $k$ roll down an inclined plane of angle $\theta$ and height $h$.
$1$. Forces acting on the body:
- Component of weight $mg \sin \theta$ acting down the plane.
- Friction force $f$ acting up the plane.
- Normal force $N$ perpendicular to the plane.
$2$. Equations of motion:
- For translational motion: $mg \sin \theta - f = ma$ (where $a$ is linear acceleration).
- For rotational motion about the center of mass: $\tau = I\alpha = fR$,where $I = mk^{2}$ and $\alpha = a/R$.
- Thus,$fR = (mk^{2})(a/R) \implies f = mk^{2}a/R^{2}$.
$3$. Solving for acceleration $a$:
- Substitute $f$ into the translational equation: $mg \sin \theta - mk^{2}a/R^{2} = ma$.
- $mg \sin \theta = ma(1 + k^{2}/R^{2}) \implies a = \frac{g \sin \theta}{1 + k^{2}/R^{2}}$.
$4$. Finding velocity $v$:
- Using $v^{2} = u^{2} + 2as$,where $u = 0$ and $s = h/\sin \theta$:
- $v^{2} = 2 \left( \frac{g \sin \theta}{1 + k^{2}/R^{2}} \right) \left( \frac{h}{\sin \theta} \right)$.
- $v^{2} = \frac{2gh}{1 + k^{2}/R^{2}}$.
Thus,the result is proved.