Power Question: Power Calculation for Particle Moving in Circular Path

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A particle of mass 'm' moves in a circular path with centripetal acceleration defined as ac = k2rt2. The power delivered to the particle by the acting force was calculated, leading to various options, with the correct answer determined to be mk2r2t. The discussion emphasized the importance of considering the force component along the particle's displacement to accurately compute power. Additionally, there was a clarification that the force acting on the particle is not solely centripetal, highlighting the need to analyze the force's direction relative to the particle's velocity. The conversation concluded with a realization of a misinterpretation of the original question regarding the nature of the force involved.
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Power Question !Need Help

Homework Statement


A particle of mass 'm' is moving in a Circular path of constant radius 'r' such that it's centripetal acceleration is given by ac= k2rt2 where k is constant. The power delivered to the particle by the force acting on it is :-

A) 2(pi)mk2r2t
B) mk2r2t
C) (mk4r2t5)/3
D) zero

Homework Equations





The Attempt at a Solution


Centripetal force will be f=mac so f=mk2rt2.

Now, Centripetal acceleration = v2/r.
so v= krt
now,
power = force*distance/time for which work is done
(mk2rt2*vt)/t

from this i am getting power as mk3r2t3.
 
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What direction does the force act? What direction is the displacement of the particle?
 


Force is centripetal so i think it is acting towards centre. And it is moving in circular track.
Do i have to take the component of force acting along displacement.
 


sorry i think force is acting away from centre
 


prateek_34gem said:
Force is centripetal so i think it is acting towards centre.
Good!
And it is moving in circular track.
Do i have to take the component of force acting along displacement.
Absolutely, in order to find the work done on the particle by that force.
prateek_34gem said:
sorry i think force is acting away from centre
No, you were right the first time.
 


prateek_34gem said:
Now, Centripetal acceleration = v2/r.
so v= krt

Yes.

now,
power = force*distance/time for which work is done
(mk2rt2*vt)/t

As Doc Al said, you need to use the force component along the particle's displacement.

Is there another equation for power that you are familiar with?
 


F.scosX/t ??
 


ohk , I got the answer.
Here is how i got it :

Let us assume that the particle moves a quarter. so its displacement will be
21/2r.
Now since the force acts in the direction towards centre.
so taking the component of displacement along force :
21/2rcos45.
now using the same formula:
Power = Force*displacement /time

(mk2r2t2*21/2rcos45)/t
=mk2r2t2*r/t
=mk2r2t

so, (B) is correct option.

Thank you guys.
PF rocks!
 


I just realized that I misread the question. :redface:
prateek_34gem said:
The power delivered to the particle by the force acting on it is
I was thinking that it said "Find the power delivered by the centripetal force", which is an entirely different question. The force acting on this particle is not simply centripetal.

Nonetheless, as already stated: Find the component of the force in the direction of the particle's velocity. Or find the energy of the particle as a function of time.
 
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