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 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.
Solution
A body rolling on an inclined plane of height h,is shown in figure below:
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
=
I ω2 +
mv2
But I = mk2 and ω = v / R
∴ Eb =
(mk2)
+ 
mv2
=
mv2 (1 +
)
From the law of conservation of energy, we have
ET = Eb
mgh =
mv2 (1 +
)
∴ v =
Hence, the result.
Let,
m = Mass of the body
R = Radius of the body m = Mass 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
=
I ω2 + mv2But I = mk2 and ω = v / R
∴ Eb =
(mk2) + mv2=
mv2 (1 +)From the law of conservation of energy, we have
ET = Eb
mgh =
mv2 (1 +)∴ v =
Hence, the result.
