# A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity of 2√2 rad/s. The radius of the cylinder must be ?

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[Take g=10 m/s^2]

a. 5 cm , b. 0.5 m, c.√10 cm , d.√5 m(answer), e.10 cm

How do you get this answer? Thanks!

[Take g=10 m/s^2]

a. 5 cm , b. 0.5 m, c.√10 cm , d.√5 m(answer), e.10 cm

How do you get this answer? Thanks!

##### 1 Answer

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

If

Moment of Inertia around the central axis

Also given is

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

Total kinetic energy is

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

#mgh=1/2mv^2+1/2Iomega^2# ......(5)

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

Solving for

Inserting given values we get value of radius