# 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!

[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!

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

#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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The radius of the cylinder is 1 meter.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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