Systems of Particles and Rotational Motion

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Question : 27
Total: 33
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
v2=
2gh
(1+k2R2)

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:  
Let M, R, k be the mass, radius and radius of gyration of a body placed at the top A of the inclined plane of height h and angle of inclination θ.
∴ Its total energy at point
A=P.E.=mgh
Its total energy at point
B = K.E. of translation + K.E. of rotation
=
1
2
m
v2
+
1
2
I
ω2

According to the principle of conservation of energy we have
1
2
m
v2
+
1
2
(mk2)
v2
R2
=mgh[I=mk2,ω=
v
R
]
1
2
m
v2
(1+
k2
R2
)
=mgh

v2=
2gh
1+
k2
R2
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