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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+k2/R2) 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.
Asked by Abhay Last Modified
Answer 
A body rolling on an inclined plane of height h,is shown in the following figure:
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, E1= mgh
Total energy at the bottom of the plane, Eb = KErot + KEtrans
= (1/2) I ω2 + (1/2) mv2
But I = mk2 and ω = v / R
∴ Eb = (1/2)(mk2)(v2/R2) + (1/2)mv2
= (1/2)mv2 (1 + k2 / R2)
From the law of conservation of energy, we have:
ET = Eb
mgh = (1/2)mv2 (1 + k2 / R2)
∴ v = 2gh / (1 + k2 / R2)
Hence, the given result is proved.
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