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Say you've tied a string to a bucket handle and you're swinging the bucket around. It has translational kinetic energy, of course, but it is also rotating (at the same angular velocity as the rotation of the string about its pivot).

Is this rotation considered when doing energy calculations? Is it ignored because there is no rotation relative to the translational velocity?

Thanks all.

EDIT: my use of the word ‘pivot’ above seems to be causing some confusion. I mean the centre of the bucket’s circular motion, i.e. the point where one end of the string is fixed. I probably should have recognised the improper wording; sorry about that.

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  • $\begingroup$ How do you find the translational kinetic energy for the bucket? $\endgroup$
    – nasu
    Commented Apr 28, 2021 at 5:00
  • $\begingroup$ @nasu Just the usual 0.5 * m * v^2. Using rotational velocity o and length of string l, you can do 0.5 * m * (ol)^2. $\endgroup$
    – ajzcole
    Commented Apr 28, 2021 at 5:09
  • $\begingroup$ What is v? Different points on the bucket have different velocities. $\endgroup$
    – nasu
    Commented Apr 28, 2021 at 5:14
  • $\begingroup$ @ajzcole can you please specify your doubt. I mean is the bucket rotating about its own axis or do you mean it is rotating in circular motion along with string. $\endgroup$
    – Dheeraj Kumar
    Commented Apr 28, 2021 at 10:57
  • $\begingroup$ @nasu I hadn’t thought about this. I intended v to be the velocity of the centre of mass of the bucket. $\endgroup$
    – ajzcole
    Commented Apr 29, 2021 at 7:11

2 Answers 2

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If you swing an object such as a bucket on a string, it undergoes both circular and rotational motion, and you should take both into account when performing energy calculations.

By circular motion, I mean motion about the centre of the circle. By rotational motion, I mean motion around the bucket's centre of mass.

If the energy due to the rotational motion of the object is very small compared with the energy due to the circular motion, then you can ignore it. That would be the case if the object is point like, or if the tangential velocity is very large compared with the angular velocity (ie if the string is very long).

If you are struggling to understand the difference, consider the following example. Imagine two identical discs of mass 1kg each glued on a circular table, 50cm from the centre of the table. If the table is made to spin about its centre, the discs will each move in the same way and will each have the same kinetic energy. Now, imagine that only one disc is glued to the table, and that the other sits on a frictionless axle sticking up from the table top. The movement of the two discs will no longer be the same. Both discs will follow a circular path around the centre of the table, but the glued disc will also rotate about its centre of mass, while the disc on the axle will not. As a consequence, the energy of the glued disc will be higher than the disc that is free to maintain its original orientation.

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You must consider both motions translating and rotation about the center of mass
for translatory motion use $\frac{mv^2}{2}$
for rotation about center of mass use $\frac{I{\omega}^2}{2}$
the sum of both energies gives the total kinetic energy of the bucket
Note: I took the bucket as a point object just for understanding the concept but actually we need to know the dimensions of the bucket for solving the question.

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  • $\begingroup$ But there is no unique v. Velocity changes from point to point on the bucket. $\endgroup$
    – nasu
    Commented Apr 28, 2021 at 5:16
  • $\begingroup$ @nasu yes, you are right but anyway the question is not about that. If you are rotating that in horizontal plane then magnitude of velocity doesn't change (assuming uniform circular motion). If you rotate in any other plane we can say that magnitude of the velocity changes, even angular velocity changes in this case. In any case, you can apply both formulas and get the correct answer. $\endgroup$
    – Dheeraj Kumar
    Commented Apr 28, 2021 at 5:24
  • $\begingroup$ It's not about velocity changing in time. At any given instant, the velocity changes from point to point. The points closer to the rotation center move slower than the points farther away from the rotation center. The bucket is not a point-like object. $\endgroup$
    – nasu
    Commented Apr 28, 2021 at 12:20
  • $\begingroup$ @nasu yes the velocity changes from point to point. For understanding the concept I just took the bucket as a point object but it is not. If we consider it as an extended object then we need to know the geometry of the bucket for finding energy $\endgroup$
    – Dheeraj Kumar
    Commented Apr 28, 2021 at 12:28

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