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.[/color]

 

[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}<br /> \vec r &amp;= r\hat r \\<br /> \vec v &amp;= r\omega\hat{\theta}\\<br /> \vec a &amp;= -r\omega^2\hat r + r\dot{\omega}\hat\theta<br /> \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}<br /> \vec r &amp;= r\hat r \\<br /> \vec v &amp;= r\omega\hat{\theta}\\<br /> \vec a &amp;= -r\omega^2\hat r + r\dot{\omega}\hat\theta<br /> \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}<br /> \vec a &amp;= (\ddot r- r\omega^2)\hat r + (r\dot{\omega}+2\dot r \omega)\hat\theta<br /> \end{align}

I love Physics ...muah..

 

[Ratch]

Tny,

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.

You are going against a well known principle of vector analysis. See the snippet below from a book on vectors. It shows that the derivative of a vector r(t) is the tangential velocity vector of the curve r(t), regardless of its path or magnitude. And, as the snippet shows, differentiating twice gives the acceleration vector. If this is true as the snippet avers, then integrating the acceleration vector gives you the back the tangential vector.

So integrating the acceleration vector was valid in the problem, and I got the correct answer doing so. I used a constant tangential velocity and acceleration in my example to easily illustrate that principle, but it still holds for a variable velocity.

Ratch

 

[TSny]

Ratch,

All I said was that integrating the acceleration vector always yields the velocity vector. Yes, the velocity vector is always tangent to the trajectory.

But, in general, it is not correct to integrate just the magnitude of the acceleration vector to get the magnitude of the velocity. Moreover, for circular motion, it's not correct to integrate the magnitude of the centripetal component of acceleration to get the magnitude of the velocity.

 

[BruceW]

Ratch, as azizlwl has written in the post above yours, the acceleration vector is
\vec{a} = ( \ddot{r} - r \omega^2 ) \hat{r} + ( r \dot{\omega} + 2 \dot{r} \omega ) \hat{\theta}
And the centripetal acceleration is r \omega^2 So you can see that in general, the integral of the centripetal acceleration will not give the velocity.

 

[Ratch]

TSny,


From your post #32.

No, you cannot get the tangential velocity by integrating the centripetal acceleration in this problem.

From your last post.

All I said was that integrating the acceleration vector always yields the velocity vector.

So which is it?

But, in general, it is not correct to integrate just the magnitude of the acceleration vector to get the magnitude of the velocity. Moreover, for circular motion, it's not correct to integrate the magnitude of the centripetal component of acceleration to get the magnitude of the velocity.

I could have integrated the acceleration vector to get the tangential velocity vector, and then found the magnitude of that vector. The result would have been the same as integrating the magnitude of the acceleration vector. The magnitude of the tangential velocity is needed for 1/2*m*v^2. Notice that the answer I got is the same as what the problem says it is.

Ratch

 

[Ratch]

BruceW,

Ratch, as azizlwl has written in the post above yours, the acceleration vector is
a⃗=(r¨−rω2)rˆ+(rω˙+2r˙ω)θˆ

The problem statement gives the acceleration as a function of k2rt2. Where is "t" in the above vector equation?

And the centripetal acceleration is rω2 So you can see that in general, the integral of the centripetal acceleration will not give the velocity.

For this problem, the centripetal acceleration is not r*ω^2. The problem statement specifically says it is k^2*r*t^2

Ratch

 

[D H]

Ratch, centripetal acceleration is a scalar quantity. It is the component of the acceleration toward the center of curvature. The only time you can treat the scalar centripetal acceleration as an acceleration vector is when this is the only component of acceleration. That isn't the case for nonuniform circular motion. There is always a nonzero tangential acceleration in this case. Integrating centripetal acceleration is not valid in the case of nonuniform circular motion.

The problem statement gives the acceleration as a function of k2rt2. Where is "t" in the above vector equation?

It's implicit in the angular velocity. It is a function of time.

For this problem, the centripetal acceleration is not r*ω^2. The problem statement specifically says it is k^2*r*t^2

Centripetal acceleration is always r\omega^2 for circular motion about the origin. Always. Google the term "Frenet–Serret formulas".

The expressions I wrote in post #34 are valid for circular motion about the origin, uniform or nonuniform. The expressions azizlwl wrote in post #36 are, with one exception, valid for any kind of planar motion. (That one exception is if and when the trajectory hits the origin; the unit vectors \hat r and \hat{\theta} are undefined at this point.)

We are given that centripetal acceleration is rk^2t^2 in this problem. Since centripetal acceleration must always be r\omega^2, one can equate the two expressions, yielding \omega^2 = (kt)^2. This in turn means that angular velocity must be either \omega = kt or \omega = -kt.

 

[TSny]

Ratch,

I said, "No, you cannot get the tangential velocity by integrating the centripetal acceleration in this problem."

I also said, "integrating the acceleration vector always yields the velocity vector."

[In the second statement "acceleration vector" denotes (as usual) the "complete" acceleration vector which would include both the centripetal and tangential components of acceleration.]

There is no inconsistency in the two statements. In this problem, the centripetal acceleration alone does not equal the "total" acceleration because there is also a tangential component of acceleration in this problem. You must integrate the complete acceleration vector in order to get the velocity vector.

You seem to be implying that in this problem you can get the velocity vector by just integrating part of the acceleration vector (namely, the centripetal acceleration alone) and leaving out the tangential acceleration. But the tangential component of acceleration is essential. It is responsible for changing the magnitude of the velocity.

 

[Ratch]

D H,

Ratch, centripetal acceleration is a scalar quantity.

Never, any acceleration has a direction and a magnitude. That makes it a spatial vector.

The only time you can treat the scalar centripetal acceleration as an acceleration vector is when this is the only component of acceleration.

That makes not sense because acceleration is never scalar. It always has a direction.

That isn't the case for nonuniform circular motion. There is always a nonzero tangential acceleration in this case. Integrating centripetal acceleration is not valid in the case of nonuniform circular motion.

The book snippet I posted shows that acceleration can be calculated no matter what the tangential velocity. For a circle with a constant r, the center of curvature never changes either.

Centripetal acceleration is always rω2 for circular motion about the origin. Always. Google the term "Frenet–Serret formulas".

Agreed, I am familar with those formulas.

We are given that centripetal acceleration is rk2t2 in this problem. Since centripetal acceleration must always be rω2, one can equate the two expressions, yielding ω2=(kt)2. This in turn means that angular velocity must be either ω=kt or

I understand your derivation which doesn't agree with mine. I am working to resolve this dichotomy.

Ratch

 

[Ratch]

TSy,

I said, "No, you cannot get the tangential velocity by integrating the centripetal acceleration in this problem."

I also said, "integrating the acceleration vector always yields the velocity vector."

Thank you for the clarification.

There is no inconsistency in the two statements. In this problem, the centripetal acceleration alone does not equal the "total" acceleration because there is also a tangential component of acceleration in this problem. You must integrate the complete acceleration vector in order to get the velocity vector.

You seem to be implying that in this problem you can get the velocity vector by just integrating part of the acceleration vector (namely, the centripetal acceleration alone) and leaving out the tangential acceleration. But the tangential component of acceleration is essential. It is responsible for changing the magnitude of the velocity.

You might be onto something. At first thought, I kind of agree with you, but I am going to have to think about it some more.

Ratch

 

[azizlwl]

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

I love Physics ...muah..

adding more

\begin{align}<br /> \vec v &amp;= (\dot r)\hat r + (r\omega)\hat\theta<br /> \end{align}

\begin{align}<br /> \hat r &amp;= i Cos\theta +j Sin\theta<br /> \end{align}
\begin{align}<br /> \hat \theta &amp;= - i Sin \theta +j Cos\theta<br /> \end{align}

 

[ehild]

I could have integrated the acceleration vector to get the tangential velocity vector, and then found the magnitude of that vector. The result would have been the same as integrating the magnitude of the acceleration vector.

Ratch

That is utterly wrong. The magnitude of an integral is not the same as the integral of the magnitude.

ehild

 

[micromass]

hjgb, which book is this problem from??

 

[hqjb]

hjgb, which book is this problem from??

1000 Solved Problems in Classical Physics Ahmad A. Kamal.

Sorry for the confusion guys, think the question is wrong, not the first time in this book :\

 

[Yukoel]

hjgb, which book is this problem from??

Hello micromass ,
As far as I know this problem appeared in the entrance test for the undergraduate entrance examination in the country India conducted by the Indian Institute of technology back in the year 1994 if my memory serves me correctly.The answer to (i suppose it was set in an objective format) is the same as shown in the attempt by hqjb.I still need an Indian to confirm it though.
That aside I suppose the thread below discusses the same problem.
https://www.physicsforums.com/showthread.php?t=257765
regards
Yukoel

 

[azizlwl]

1000 Solved Problems in Classical Physics Ahmad A. Kamal.

Sorry for the confusion guys, think the question is wrong, not the first time in this book :\

http://img402.imageshack.us/img402/2748/soklan.jpg

The answer from the book.
http://img151.imageshack.us/img151/1207/salahq.jpg