Question

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 ?

[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!

Answer 1

When the solid cylinder rolls down the inclined plane, without slipping, its total kinetic energy is given by
`KE "due to translation"+"Rotational " KE="1/2mv^2+1/2Iomega^2` ....(1)

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If `r` is the radius of cylinder,
Moment of Inertia around the central axis `I=1/2mr^2` .....(2)
Also given is `omega=v/r` ....(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=1/2mv^2+1/2Iomega^2` ......(5)

Using (3) and (4), equation (5) becomes
`mgh=1/2m(romega)^2+1/2xx1/2mr^2omega^2`
`=>gh=(1/2+1/4)r^2omega^2`
`=>gh=3/4r^2omega^2`
Solving for `r`
`=>r=sqrt((4gh)/(3omega^2))`
Inserting given values we get value of radius `r` as

`r=sqrt((cancel4xx10xxcancel3)/(cancel3(cancel2sqrt2)^2)`
`r=sqrt5m`