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A particle of mass m is moving in a circ...

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c) = k^(2)rt^(2)`, where k is a constant. The power delivered to the particle by the forces acting on it is :

A

`2pi m K^(2) r^(2) t`

B

` m K^(2) r^(2) t `

C

` (mK^(4)r^(2) t^(5))/( 3) `

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
For U.C.M ., centripetal acceleration ` a_(c ) =(v^(2))/(r )`
But it is given that `a_(c ) ` is variable .hence the particle has both `a_(c ) and a_(T ) ` ( tangential acceleration ) .
Given : `a_(c ) = K^(2) r t ^(2) = (v^(2))/(r ) " " therefore v^(2) =K^(2) r^(2) t^(2) therefore V = Krt `
` and a_(T ) =(dv )/(dt ) =(d )/(dt ) (krt) = Kr and ` Force `ma_(T )= mkr `
Power is required for tangential acceleration
For `A_(c ) , W = 0 and P=0`
` therefore `Power = Force `xx` Velocity
`therefore P= mKr xxKrt = m K^(2)r^(2) t`
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