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

[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

 

[Dadface]

The book answer above uses the equation:

P=Fv where F is the centripetal force

This equation can be derived from:

Work = force * distance=Fdx

Power=Work/time =Fdx/dt

With many problems we can write dx/dt=v but I think it is incorrect to do so in this case because the radius remains constant and dx remains at zero.

 

[Infinitum]

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

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

I think there is a mistake here. Power is given as a dot-product,

P = F\cdot v

I don't know how they are assumed to be in the same direction.

 

[TSny]

Perhaps the problem meant to say that the tangential acceleration varies as ...

Then the answer stated in the problem would be correct.

 

[Infinitum]

Perhaps the problem meant to say that the tangential acceleration varies as ...

Then the answer stated in the problem would be correct.

If that is the case, then it would definitely be correct

 

[D H]

Perhaps the problem meant to say that the tangential acceleration varies as ...

Then the answer stated in the problem would be correct.

Nice explanation, but wrong. The Indian education system apparently works (at least in part) by teaching the test. Old tests, to be precise. Perhaps India's copyright laws make those old tests a quick and easy source of homework questions for authors of Indian textbooks. In any case, this question is replicated over and over in many Indian physics texts. See the following google books search:

http://www.google.com/search?q=A+pa...ius+r+such+that+its+centripetal+acceleration"

Many just keep the question in its original form, a multiple choice question with choices 2πmk²r²t, mk⁴r²t⁵/3, mk²r²t, and zero. Some change the choices, others make the question open ended (not multiple choice). A small number of those books provide solutions -- including some erroneous ones (e.g., the book that is the subject of this thread). Apparently it is easy for authors to get their hands on the old test questions, but not on the old test answers.

 

[Saitama]

Nice explanation, but wrong. The Indian education system apparently works (at least in part) by teaching the test. Old tests, to be precise. Perhaps India's copyright laws make those old tests a quick and easy source of homework questions for authors of Indian textbooks. In any case, this question is replicated over and over in many Indian physics texts. See the following google books search:

http://www.google.com/search?q=A+pa...ius+r+such+that+its+centripetal+acceleration"

Many just keep the question in its original form, a multiple choice question with choices 2πmk²r²t, mk⁴r²t⁵/3, mk²r²t, and zero. Some change the choices, others make the question open ended (not multiple choice). A small number of those books provide solutions -- including some erroneous ones (e.g., the book that is the subject of this thread). Apparently it is easy for authors to get their hands on the old test questions, but not on the old test answers.

A good book is present in the link you gave D H, the book is "The Pearson Guide to Objective Physics for the IIT-JEE 2012". You can see the question in the preview and the answer is given the same as OP. Scroll down a few pages after the question and you will see the answer key.
Click Me

I think there is a mistake here. Power is given as a dot-product,
P=F⋅v
I don't know how they are assumed to be in the same direction.

Seems like the author did the same mistake as me.

 

[Ratch]

D H,

It looks like TSny found the real reason why my method did not work.

TSny,

Thanks for your insight.

ehild,

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

You are correct, I should have known better.

Ratch