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when we talk about circular motion we have an equation of $v=r\omega$ where $v$ is linear velocity $r$ is the radius of circular motion and $\omega$ is its angular velocity but if circular motion has a constant angular velocity it means that both $\omega$ and $r$ are constant so acc to this equation $v=r\omega$ linear velocity should also be constant. The same goes for acceleration $a=r\alpha$ if $\omega$ is constant $\alpha$ is zero and so much a but as velocity is changing a can not be zero. How can we explain this paradox? I have just started my high school so I am aware of only simple calculus so I will appreciate if you take this in your consideration while answering :) thank you

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    $\begingroup$ It is likely that you are confusing between an object rotating along an axis vs an object rolling. The v here has different meanings $\endgroup$ May 3 '20 at 17:20
  • $\begingroup$ If your system has an angular acceleration, α, then angular velocity, ω, and the tangential velocity are not constant. $\endgroup$
    – R.W. Bird
    May 3 '20 at 18:12
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In $v=\omega r$, $v$ is speed, not velocity. The speed stays constant in uniform circular motion.

The vector relationship involving velocity for uniform circular motion is $\vec{v}=\vec{\omega}\times\vec{r}$. Both $\vec r$ and $\vec v$ change with time, but their magnitudes $r$ and $v$ don’t.

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  • $\begingroup$ I suppose, $v$ is velocity, but the formula looks like $\vec v=\vec \omega \times \vec r$? $\endgroup$ May 3 '20 at 17:23
  • $\begingroup$ @user12986714 Yes. I was adding that as you were commenting. $\endgroup$
    – G. Smith
    May 3 '20 at 17:27
  • $\begingroup$ For the above vector (or cross) product, the vector representing the ω is defined as being along the axis of rotation ( the only direction in a rotating system which does not change with time). (Wrap your right hand fingers around the axis in the direction of rotation. Then your thumb defines the direction for the w vector. The r vector goes from the axis to a point in the object which is moving in a circle. The product (the velocity vector) is at a right angle to both of these in a direction determined by a another right hand rule. In this case tangent to the circle. $\endgroup$
    – R.W. Bird
    May 3 '20 at 17:58
  • $\begingroup$ In general to find the direction of a vector product, hold your right hand so that you can curl your fingers from the direction of first named vector in the product to the direction of the second named vector. Your thumb gives the direction of the product. $\endgroup$
    – R.W. Bird
    May 3 '20 at 18:04
  • $\begingroup$ Thank you for your answer. I highly appreciate it. I just leaned this topic yesterday and my teacher did not tell me this exact equation. But I was so curious to know I could not wait to ask him tomorrow. $\endgroup$ May 3 '20 at 19:06
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In the equation $v=r \omega$, as used at high school level, all three quantities are scalars. In particular, $v$ is not velocity but the magnitude of the velocity, that is $speed$.

For a body in circular motion at constant speed, $r$, $v$ and $\omega$ are all constant. But the body's velocity is constantly changing because the direction in which it is moving is constantly changing. By analysing this in detail we find that the body has an acceleration towards the centre of the circle. There is no tangential acceleration, that is acceleration 'around' the circle.

The equation $a=r \alpha$ that you quote is again a relationship between scalars. $a$ in this equation is the magnitude of the tangential component of a body's acceleration. So it is zero for a body moving in a circle at constant speed. Only if the body's speed is changing is $a$ non-zero.

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Not only in a circular motion, but also in any motion, you can think the acceleration as composed by two terms:

  • one is the tangential (or linear) acceleration, that takes into account the change in linear velocity, and is parallel to the motion: $$a_t=\alpha r$$
  • the other is the centripetal acceleration, that describes the change in direction and is perpendicular to the motion: $$a_c = {v^2 \over r}=\omega^2 r $$

In particular in a uniform circular motion $\omega$ and $r$ -as you have said- are constant and $\alpha = 0$, so the one term that is zero is the tangential acceleration.

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It's a sloppy notation, the $(t)$s are usually dropped from $v(t)=r\omega(t)$, hence $v=r\omega$ can either be constant or variable. One needs to pay attention to the problem statement he has in front of him!

In any case, you are right; if $\omega$ is constant, i.e $v$ is also constant, then $\alpha$ is zero, and the same goes for $a$ since they are related by the equation you have written.

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Notice your slight mistake. You have considered 'both $\omega$ and R as constants. But, if you think about it, the $\vec r$ is also rotating. So, the velocity is not constant, as you had interpreted. Now, coming to the acceleration, (using the same arguement) it too is not zero.

enter image description here

As you can obviously see from the image, $\vec {r_1}$ and $\vec {r_2}$ are not equal, but their magnitudes are (as Mr. G. Smith has said)

Think of it like this. The particle is moving momentarily in a straight line. But it is accelerated towards the centre, so it falls towards the centre, but perpetually keeps doing this without getting any closer. For instance, if you throw a ball standing on a seashore, perfectly horizontally such that it falls beyond the horizon, due to the round shape, it'll go into perpetual motion.

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It's not perfectly true as the exact relationship is $$\vec{v}=\vec{\omega}\times\vec{r}$$

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    $\begingroup$ The OP’s equation is perfectly true. The OP just didn’t understand what the $v$ in it means. (But I didn’t downvote.) $\endgroup$
    – G. Smith
    May 3 '20 at 17:30

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