Question

A cylinder of radius 4cm and mass 2kg in rolling on a horizontal surface with a velocity of 0.2m/s . calculate its kinetic energy, for a cylinder I=MR^2/2.

Answer 1

Answer:

When the solid cylinder rolls down the inclined plane, without slipping, its total kinetic energy is given by 

KEdue to translation+Rotational KE=12mv2+12Iω2....

If r is the radius of cylinder, 

Moment of Inertia around the central axis I=12mr2 .....(2)

Also given is ω=vr ....(3)

Assuming that it starts from rest and ignoring frictional losses, at the bottom of the plane 

Total kinetic energy is =Potential Energy at the top of plane=mgh.....(4)

Using Law of conservation of energy and equating (1) and (4) we get

mgh=12mv2+12Iω2 ......(5)

Using (3) and (4), equation (5) becomes 

mgh=12m(rω)2+12×12mr2ω2

⇒gh=(12+14)r2ω2

⇒gh=34r2ω2

Solving for r

⇒r=√4gh3ω2

Inserting given values we get value of radius r as

r=⎷4×10×33(2√2)2

r=√5m