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Question
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 = (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.
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Solution 1
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, E1= mgh
Total energy at the bottom of the plane, `E_b = KE_rot + KE_trans`
`=1/2 Iomega^2 + 1/2 mv^2`
But `I = mk^2 " and " omega = v/r`
`:.E_b = 1/2 (mk^2)(v^2/R_2) + 1/2 mv^2`
`=1/2 mv^2 k^2/R^2 + 1/2mv^2`
`= 1/2 mv^2(1+ k^2/R^2)`
From the law of conservation of energy, we have:
`E_T = E_b`
`mgh = 1/2mv^2(1+k^2/R^2)`
`:.v = (2gh)/(1+k^2"/"R^2)`
Hence, the given result is proved.
Solution 2
Let a rolling body (I = Mk2) rolls down an inclined plane with an initial velocity u = 0; When it reaches the bottom of the inclined plane, let its linear velocity be v. Then from conservation of mechanical energy, we have Loss in P.E. = Gain in translational K.E. + Gain in rotational K.E.
`Mgh = 1/2mv^2 + 1/2 Iomega^2`
`= 1/2mv^2 + 1/2(mk^2)(v^2/R^2)`
`Mgh = 1/2mv^2 (1+k^2/R^2)`
`v^2 = (2gh)/(1+k^2/R^2)`
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