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 .

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 .Question 7.27 .System Of Particles And Rotational Motion

Answer 1

Let m = mass of rolling body
R = radius and K = radius of gyration of rolling body on inclined plane of height h .

At highest point ,
energy of body (Ki)= PE = mgh
Let v and w be the linear and angular velocity of body at bottom of inclined plane respectively.

At lowest point ,
Energy of body(Ef) = linear kinetic energy + rotation kinetic energy
= 1/2 × mv² + 1/2× Iw²
We know,
Moment of inertia of body (I) = mK² use this here,

Ef = 1/2 × mv² + 1/2×mK²w²

A/c to law of conservation of energy,
Ei = Ef
mgh = 1/2m( v² + w²K²)
2gh = v² + K²w²
We also know,
w = v/R use this above ,
2gh = v² + K²v²/R²
2gh = v²( 1 + K²/R²)
v² = 2gh/( 1 + K²/R²)
Hence, proved

Answer 2

Answer:

Let a rolling body (I = Mk2) rolls down an inclined plane with an initial velocity u = 0; When it reaches the bottom of 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.