Question

Derive expressions for the final velocity and the total energy of a body rolling without slipping.

Answer 1

Let us consider a rigid body of mass M and radius R rolling down inclined plane from a height h. Let v be the linear velocity acquired by the body when it reaches the bottom of the plane and k is its radius of gyration.

From law of conservation of energy theorem,
potential energy at the top of inclined plane = kinetic energy at the bottom of inclined plane.

or, potential energy at the top =translational K.E + rotational K.E

e.g., Mgh = 1/2 Mv² + 1/2 Iw²

Or, Mgh = 1/2Mv² + 1/2 Mk²(v²/R²) [ as I = Mk² and w = v/R ]

or, Mgh = 1/2 Mv²[1 + k²/R² ]

So, v = √{2gh/(1 + k²/R²)}

Hence, expression of final velocity is v = √{2gh/(1 + k²/R²)}

And total total energy = translational kinetic energy + rotational kinetic energy
= 1/2Mv² + 1/2Iw²
= 1/2Mv² + 1/2Mk² (v²/R²)
= 1/2Mv²[1 + k²/R²]

Answer 2

Answer:

Given,

moment of inertia = 0.3 kgm²

angular velocity = 300 rev/min = 300 × 2π/60 rad/s = 10π rad/s

Use formula, \omega=\omega_0+\alpha tω=ω

0

+αt

0 = 10π + \alphaα × 20

[ Final angular velocity is taken zero because body becomes rest after 20sec.]

Angular acceleration, \alphaα = -π/2 rad/s

We know, Torque = moment of inertia × angular acceleration.

so, torque = 0.3 × -π/2 = -0.15π Nm.

hence, magnitude of torque is 0.15π Nm.