### Q.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} = 2*gh* / [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.

**Answer:-**

A body rolling on an inclined plane of height *h*,is shown in the following figure:

*m*= Mass of the body

*R *= Radius of the body

*K* = Radius of gyration of the body

*v *= Translational velocity of the body

*h *=Height of the inclined plane

*g* = Acceleration due to gravity

Total energy at the top of the plane, *E*_{1}=* m*g*h*

Total energy at the bottom of the plane, *E*_{b} = KE_{rot} + KE_{trans}

= (1/2) *I* ω^{2} + (1/2) *mv*^{2}

But *I* = *mk*^{2} and ω = *v* / *R*

∴ *E*_{b} = (1/2)(*mk*^{2})(*v*^{2}/*R*^{2}) + (1/2)*mv*^{2}

= (1/2)*mv*^{2} (1 + *k*^{2} / *R*^{2})

From the law of conservation of energy, we have:

*E*_{T} = *E*_{b}

*mgh* = (1/2)*mv*^{2} (1 + *k*^{2} / *R*^{2})

∴ *v* = 2*gh* / (1 + *k*^{2} / *R*^{2})

Hence, the given result is proved.