__Question 7.27__:*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}= 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.

__Solution__:*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) m v

^{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 = 2gh / (1 + k

^{2}/ R

^{2})

Hence, the given result is proved.

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