Circular Motion Power: Why Isn't Work Done Equal to Zero?

[hqjb]

1. Homework Statement [/b]

A particle of mass m is moving in a circular path of constant radius r such
that its centripetal acceleration a varies with time t as $a = k^2rt^2$, where k is a
constant. Show that the power delivered to the particle by the forces acting on
it is $mk^4r^2t^5/3$

Homework Equations

The Attempt at a Solution

Why isn't work done, hence power = 0 since it's moving in a circle and resultant force is centripetal?

Edit : Assuming there's a tangential component,

I take
$a_c = k^2rt^2 = v_{tan}^2/r$
$v_{tan} = √k^2r^2t^2 = krt$
$a_{tan} = dv/dt = kr$
$P = Fv = ma_{tan}v_{tan} = krt * kr * m ≠ mk^4r^2t^5/3$

 

[Saitama]

I have got the answer you have mentioned but i am not sure what could be the reason that my answer is correct. I am sitting on my computer desk and following your thread but no one is yet replying, so i think i will post my solution and see if i could receive any feedback. I know its your thread hqjb, i hope you don't mind.

I integrated the expression for centripetal acceleration with respect to time and found $v=\frac{k^2rt^3}{3}$, (now i don't know which velocity is this. )

P=Fv
P=ma_cv
$P=m*k^2rt^2*\frac{k^2rt^3}{3}$
$P=\frac{mk^4r^2t^5}{3}$

But still i don't have any reasoning to why is this correct?

 

[azizlwl]

a=k²rt²
v=k²rt³/3

E=1/2mv²=1/2m(k²rt³/3)²

P=dE/dt=(6/18)m(k⁴r²t⁵)

 

[hqjb]

a=k²rt²
v=k²rt³/3

E=1/2mv²=1/2m(k²rt³/3)²

P=dE/dt=(6/18)m(k⁴r²t⁵)

I get both of your workings, but I'm not sure why both of you used the radial component of velocity? (integrating centripetal acc = centripetal vel. right?)

Shouldn't only tangential component count in doing work?

 

[azizlwl]

I'm looking for the instantaneous velocity from instantaneous acceleration,irrespective of direction.
Energy is scalar.

 

[takudo_1912]

I think that:
v$^{2}$=v$^{2}_{c}$+$^{2}_{t}$
But v$_{t}$=krt
And v$_{c}$=$\frac{k^{2}.r.t^{3}}{3}$
W=1/2.m.$v^{2}$
P=dW/dt=...

 

[hqjb]

I think that:
v$^{2}$=v$^{2}_{c}$+$^{2}_{t}$
But v$_{t}$=krt
And v$_{c}$=$\frac{k^{2}.r.t^{3}}{3}$
W=1/2.m.$v^{2}$
P=dW/dt=...

I get my original answer + everyone else's answer in this case :

$k^2r^2tm + (k^4r^2t^5m)/3$

 

[takudo_1912]

I get my original answer + everyone else's answer in this case :

$k^2r^2tm + (k^4r^2t^5m)/3$

I think this was the correct answer :-s

 

[hqjb]

Okay after thinking about it again, I think by definition circular motion means there is no "radial velocity". My bad for that mistake. But can anyone tell me what you get when you integrate radial acceleration? What about the v^2/r formula then(My notes says that it works for non-uniform circular motion)?

 

[ehild]

I take
$a_c = k^2rt^2 = v_{tan}^2/r$
$v_{tan} = √k^2r^2t^2 = krt$
$a_{tan} = dv/dt = kr$
$P = Fv = ma_{tan}v_{tan} = krt * kr * m ≠ mk^4r^2t^5/3$

Your result looks fine. The centripetal acceleration is v²/r, even in case of non-uniform circular motion. There is no work done by the centripetal force if the particle moves along a circle. But there should be some tangential force to ensure increase of speed. The tangential force multiplied by the velocity (which is tangential of course) gives the power.

ehild

 

[azizlwl]

My understanding is when a body moves at
a=Ct² where C=k²r
The velocity will be
v=Ct³/3

KE=1/2mv²

 

[ehild]

The particle moves along a circle with constant radius r

The velocity is tangent to the path.
The tangent of the circle is perpendicular to its radius : there is no "radial velocity".

There is radial acceleration: it is called "centripetal acceleration". There can be also tangential acceleration.
The centripetal force does not do any work.
The integral of the centripetal acceleration is neither velocity nor speed.
Think of the uniform circular motion. The centripetal acceleration is constant. Does the speed increase linearly with time?

ehild

 

[Ratch]

ehild,

Your result looks fine. The centripetal acceleration is v2/r, even in case of non-uniform circular motion.

I don't think so. Let ω be constant and vector R = cosωt*i + sinωt*j. Then velocity vector V = dR/dt = -ω*sinωt*i + ω*cosωt*j , where V and R are at right angles to each other. So the acceleration vector is A = dV/dt = -ω^2(cosω*t*i + sinωt*j) = -ω^2*R, which is a vector in the opposite direction of R and whose magnitude is |V|^2/|R|. This derivation requires that ω be constant (uniform). Besides, the problem statement says that the centripetal acceleration is k^2*r*t^2.

Ratch

 

[azizlwl]

Really confusing for me
If v≠v'.
Then Δv/v≠ Δs/r


http://img213.imageshack.us/img213/6705/circularmotion.jpg

But from wiki.
http://en.wikipedia.org/wiki/Non-uniform_circular_motion
Radial acceleration is always equal to v²/r

 

[ehild]

Really confusing for me
If v≠v'.
Then Δv/v≠ Δs/r

But from wiki.
http://en.wikipedia.org/wiki/Non-uniform_circular_motion
Radial acceleration is always equal to v²/r

Read
http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration

ehild

 

[Saitama]

@ehild: Can i get some feedback on my solution? That would be very helpful. The solution is posted on page 1, post #2.

 

[ehild]

@ehild: Can i get some feedback on my solution? That would be very helpful. The solution is posted on page 1, post #2.

Pranav, think of the uniform circular motion. The centripetal acceleration is constant. Do you get anything reasonable if you integrate it with respect time? Is it the velocity or speed or anything? Is the speed of the uniform circular motion a linear function of time instead of being constant?

Read my post #13.

ehild

 

[ehild]

The same thing as the diagram your link shows. Specifically, A = the vector sum of Ac and At.

Ratch

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

 

[Saitama]

Is the speed of the uniform circular motion a linear function of time instead of being constant?
ehild

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]

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.

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?

I really have no idea what we get if we integrate the centripetal acceleration.

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]

Nice post, Yukoel!

ehild

Thanks ehild
regards
Yukoel

 

[Saitama]

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?

Thanks for the explanation ehild!

Integrating centripetal acceleration really doesn't make any sense.

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

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]

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)?.

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?

 

[Saitama]

...Descartes components...
ehild

Never heard of it, any link would help.

 

[ehild]

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

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,
$$\begin{align} \vec r &= r\hat r \\ \vec v &= r\omega\hat{\theta}\\ \vec a &= -r\omega^2\hat r + r\dot{\omega}\hat\theta \end{align}$$
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 $r\omega^2$. Power is given by $\vec F\cdot \vec v = m\vec a \cdot \vec v = mr^2\omega\dot{\omega}$ for circular motion. Here we are given that centripetal acceleration is $rk^2t^2$, making $\omega = \pm kt$. With this result, the power becomes $P=mr^2k^2t$.

 

[azizlwl]

One last note, my solution. For circular motion, uniform or not,
$$\begin{align} \vec r &= r\hat r \\ \vec v &= r\omega\hat{\theta}\\ \vec a &= -r\omega^2\hat r + r\dot{\omega}\hat\theta \end{align}$$
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 $r\omega^2$. Power is given by $\vec F\cdot \vec v = m\vec a \cdot \vec v = mr^2\omega\dot{\omega}$ for circular motion. Here we are given that centripetal acceleration is $rk^2t^2$, making $\omega = \pm kt$. With this result, the power becomes $P=mr^2k^2t$.

May I add.
For an object moving in a curved path in a plane.
$$\begin{align} \vec a &= (\ddot r- r\omega^2)\hat r + (r\dot{\omega}+2\dot r \omega)\hat\theta \end{align}$$

I love Physics ...muah..