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Does centripetal acceleration deliver power to the revolving object?

What I think:

Power is given by F.v Since for a body in circular motion velocity and acceleration are at 90 degrees the power delivered should be zero.

Does changing(increasing or decreasing)centripetal force deliver power to a revolving body?

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Assuming circular motion (like you are), the force (and acceleration) are pointed towards the centre of the circle, while the velocity is tangential, which indeed causes the magnitude of the power, $|\vec{F}\cdot\vec{v}|=|\vec{F}||\vec{v}|\cos{\theta}$ to vanish, because $\theta=\pi/2$ radians.

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What if centripetal force is varying? – Rajath Krishna R Sep 14 '13 at 15:18

The answer given by @Danu is correct. This is an elaboration of your case where the centripetal force varies.

Two things can occur when the centripetal force varies:

Case 1:

The tangential velocity varies by an appropriate amount to maintain circular motion.
Basically $$F_{\mathrm{centripetal}}=m\frac{v^2}r$$ should satisfy with the new force and velocity.
If this is the case, the power delivered is still $0$, as the force is still perpendicular to the velocity.

Case 2:

Velocity doesn't vary by an appropriate amount. So $$F_{\mathrm{centripetal}}\neq m\frac{v^2}r$$ This will cause the object to obtain a non-zero radial component of velocity. The motion will no longer be circular, and the velocity will no longer be perpendicular to force.
In this case power will be given by $P=\vec F\cdot\vec v$, which can be simplified to $P=Fv_{\mathrm{radial}}$

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Note that the velocity can be varied in Case 1 by a tangential impulse(if change in F is sudden) or by an appropriate tangential force(if change in F is continuous) – udiboy1209 Sep 14 '13 at 16:23

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