Question

Prove the result that the velocity v of a 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 $V^2=\frac {2gh}{[1 +(\frac{k^2}{R^2})]}$ 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.

Answer 1

It is given that , a body rolling on an inclined plane of height h, is shown in figure:

Mass of the body = m
Radius of the body = R
Radius of gyration of the body = K
Translational velocity of the body = v
Height of the inclined plane = h
Acceleration due to gravity = g = 9.8 m/s²

Total energy at the top of the plane = potential energy at height h from the ground = mgh
Total energy at the bottom of the plane = translational kinetic energy + rotational kinetic energy

 = 1/2 Iω² + 1/2 mv² ....…(i)
Where I is moment of inertia so, I = mK²
And ω is angular velocity, so, ω = v/R

Using in equation (i)
Total kinetic energy  = 1/2 mK²(v/R)² + 1/2 mv²
Total kinetic energy = 1/2 mv² (1 + K²/R²)

We know that total energy at the top will be equal to total energy at the bottom, by law of conservation of energy,

so, mgh = 1/2 mv²(1 + K²/R²)

⇒ v = 2gh/(1 + K²/R²)

Hence proved.

Answer 2

Answer:

A man stands on a rotating platform with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then bring his arms close to his body with the distance of each weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m²a. What is his new angular speed? (Neglect friction)b. Is kinetic energy conserved in the process? If not from where does the change