Circular motion power

ehild

We totally agree then. The acceleration is the vector sum of Ac and At. Ac is the centripetal acceleration. Its magnitude is v²/r.

ehild

Pranav-Arora

The speed is constant in uniform circular motion. The centripetal acceleration is constant in uniform circular motion but here the centripetal acceleration is varying with time.
I really have no idea what we get if we integrate the centripetal acceleration.

Yukoel

Hello hqjb and others,
I think the solution pasted in the attempt is correct.Circular motion uniform or non uniform “indeed requires the centripetal acceleration to be equal to(V^2/R)”.Let us prove it through calculus and polars.(2Dimensional avoiding the usage of axis of rotation)
We know that vector r of any point on the circle with O as origin at an elevation from say the horizontal (the diagram can be drawn easily) and with unit vectors as i and j is given as
R=R(cos(θ)i+sin(θ)j)
We define two other unit vectors e(r) and e(t) ,with the first pointing away from the center towards the radius and the second in the direction of the advancing tangent.
e( r )= cos(θ )i+sin(θ )j
e( t )= -sin(θ )i+cos(θ )j
e’( r )= e(t)*θ’(Differentiation w.r.t any variable)
Similarly
e’( t)=-e( r )* θ’
As such r=r*e( r ) ...........$
V=rω*e(t)+r’e(r) ………………$$
/* Unless radius is changing w.r.t time velocity is always tangential as r’=0 */so
V=rω*e(t) ………………………$$$
(ω is angular velocity)
So that
a=-r(ω^2)e(r) +rαe(t) ………………$$$$ (radius taken constant once again)(alternatively use complex numbers for the same result)
In short only given that radius is constant the centripetal acceleration is ((v^2)/r) no matter what (we used v=rω in the radial portion of the last equation which arrives from our velocity vector expression)For other curvilinear trajectories the centripetal acceleration is defined by the general term “normal component of acceleration”The same equation works with only the r in the denominator being radius of curvature of the curve at that point .
So
(1) Integrating centripetal acceleration in any case of circular motion is not an act that brings forth any velocity.
(2) One need not worry about radial component of velocity here as it reduces to zero given the fact that radius is constant.
(3) Something is wrong with the question, it yields fine with the acceleration being defined plainly but is incorrect as regards circular motion and centripetal acceleration.
Apologies in advance for straying the discussion off topic if I did so.Correct me if I am wrong.
Regards
Yukoel

ehild

Nice post, Yukoel!

ehild

ehild

You integrated the centripetal acceleration to get the velocity thinking (I guess) that "velocity is integral of acceleration". You could do the same with constant centripetal acceleration could you not? And do you get anything that has sense?
See an example: a ball moves along a horizontal circle with radius R=1 with 2 m/s speed. So the centripetal acceleration is 4 m/s². Integrate with respect to time: it is ∫adt=4t+const. What kind of speed is it? It is not the speed of the ball, as it is 2 m/s. It is not "radial velocity" as the radius is constant. Is it anything?

You have integrated the centripetal acceleration, so you need to know the reason why you did it.

You got the integral of the centripetal acceleration. It is not related to the velocity.

ehild

Yukoel

Thanks ehild
regards
Yukoel

Pranav-Arora

Thanks for the explanation ehild!

Integrating centripetal acceleration really doesn't make any sense.

For finding out the answer as posted by OP, i tried integrating centripetal acceleration and by luck, i was able to get that answer. So the claimed answer is wrong (i guess)?.

ehild

I guess it is wrong. And I also guess they followed the same logic, that "velocity is integral of acceleration". It is true for the vectors and for the Descartes components, but not for the centripetal acceleration.

Well, getting the same answer as in the book is not always a "luck"

ehild

schaefera

Shouldn't the answer be Power=m*k^2*r^2*t?

Pranav-Arora

Never heard of it, any link would help.

ehild

It is. That is the solution obtained and showed by the OP.


ehild

schaefera

The goal of the problem was incorrect then, as stated in the original post?

schaefera

So is the power given by dE/dt or by Fv, because the value of these two seem to differ (dE/dt=k^2r^2t, but Fv is the value in the original post).

TSny

They are the same. The OP gave a correct derivation of Fv =m k²r²t. This also equals dE/dt.

The confusion is due to the fact that the statement of the problem implied that the answer for the power should be something different than mk²r²t.

TSny

No, you cannot get the tangential velocity by integrating the centripetal acceleration in this problem. Your derivation in post #35 is valid only for constant ω.

In general, integrating the total acceleration vector with respect to time will get you the total velocity vector. And that's what you did for the case of constant ω.

But the mistake that some here are making is trying to get the tangential speed by integrating only the magnitude of the centripetal acceleration. That's not valid.

Moderator note: This post refers back to a deleted post, so don't let that confuse you. I restored this post because TSny's identification of the conceptual mistake being made is spot on.

D H

If this thread looks a bit disjointed it is because some incorrect posts and responses to them have been removed.

The OP is correct, the answer in the book is not.

D H

One last note, my solution. For circular motion, uniform or not,
[tex]\begin{align}
\vec r &= r\hat r \\
\vec v &= r\omega\hat{\theta}\\
\vec a &= -r\omega^2\hat r + r\dot{\omega}\hat\theta
\end{align}[/tex]
Note that velocity is tangential; that's always the case for circular motion, uniform or not. A couple of other common features for circular motion are centripetal acceleration and power. Centripetal acceleration is always [itex]r\omega^2[/itex]. Power is given by [itex]\vec F\cdot \vec v = m\vec a \cdot \vec v = mr^2\omega\dot{\omega}[/itex] for circular motion. Here we are given that centripetal acceleration is [itex]rk^2t^2[/itex], making [itex]\omega = \pm kt[/itex]. With this result, the power becomes [itex]P=mr^2k^2t[/itex].