Question

A solid cylinder of mass m and radius r starts rolling down an inclined plane of inclination theta. Friction is enough to prevent slipping. Find the speed of its centre of mass when its centre of mass has fallen a height h.

Answer 1

The speed of the centre of mass is $\sqrt{\frac{4}{3}gh }$

A solid cylinder of mass m and radius r starts rolling down an inclined plane.

1. Since there is no slipping, this is a problem of pure rolling.
2. We have to find the velocity of the center when it has fallen a height of 'h'.
3. Initially, the cylinder is at rest. It has only potential energy.
4. When it starts rolling, the potential energy gets converted to kinetic energy. Potential energy lost = kinetic energy gained.
5. The kinetic energy is divided into two components- rotational and translational.

•      mgh = $\frac{1}{2}$ I ω² + $\frac{1}{2}$ m ν²

        Here moment of inertia of cylinder = $\frac{m.r^{2} }{2}$

        and ω = ν/r

•      mgh = $\frac{mr^{2}v^{2} } {4r^{2} }$ + $\frac{mv^{2} }{2}$    
•      4gh = 3v²
•       v =   $\sqrt{\frac{4gh}{3} }$

   

Answer 2

Given:

A solid cylinder of mass m radius r rolling down an inclined plane of inclination Ф .

To find:

speed of its centre of mass when its centre of mass has fallen a height h

Solution:

Conservation of energy

pure rolling

initial condition:

initially solid cylinder is at rest.

it only has potential energy

UE = mgh KE = 0

TE = mgh

final condition :

UE = 0 KE =  $\frac{1}{2} m v^{2} + \frac{1}{2} I w^{2}$

where I is the moment of inertia about centre of mass of solid cylinder.

w is the angular velocity.

I for a solid cylinder = $\frac{mr^{2} }{2} }$

v = wR

KE (final) = $\frac{3mv^{2} }{4}$

TE (final) = KE (final)

total energy is conserved in case of pure rolling

hence TE(initial) = TE (final)

mgh = $\frac{3mv^{2} }{4}$

v = $\sqrt{\frac{4gh}{3} }$

Speed of centre of mass after falling a height h is equal to v = $\sqrt{\frac{4gh}{3} }$