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!

enter image source here

[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!

enter image source here

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Dec 21, 2016

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)

isaacphysics.org

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#

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