If there is a device in which a ball is attached to a pivot by a rod of length R and it is left to oscillate by dragging it to one side so that it is completely horizontal, at any point on its decent the ball has some velocity and also has some angular velocity about the pivot. In this case is the rotational kinetic energy about pivot equal to translational kinetic energy of the ball ? Why/ Why not ? If they are same then what is better way to deal with the calculations of such problems, the rotational KE about the pivot or translation KE of the ball alone ?
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asked Jan 11, 2014 at 17:11
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$\begingroup$ Why don't you try to solve it? $\endgroup$– jinaweeCommented Jan 11, 2014 at 19:02
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$\begingroup$ @jinawee : See the diagram in the answer below, if we stand at point P the ball is moving, does it have translational or rotational or both KE ? And what are you aksing me to solve myself ? $\endgroup$– Rijul GuptaCommented Jan 11, 2014 at 19:48
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$\begingroup$ I mean you could include your thoughts about the kinetic energy. In your example, does the ball spin? $\endgroup$– jinaweeCommented Jan 11, 2014 at 20:01
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$\begingroup$ No, initially my thoughts were that the ball had a certain velocity while it was coming down and was rotating about the pivot at the same time, the question then became that are both these same things or different( the translational and the rotation) $\endgroup$– Rijul GuptaCommented Jan 11, 2014 at 20:14
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$\begingroup$ As I said in the answer, it depends of the origin of the axis: if you choose P, then it has rotating energy, but not translational. And if you choose ball's center mass, then it has translational but not rotational. And, if the rod mass is 0, both energies are the same. However I wanted to provide the full explanation in the answer (I don't know if I have explained it clear enough) $\endgroup$– Victor BuendíaCommented Jan 11, 2014 at 22:58
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1 Answer
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Well, let's see if I understood correctly the question. You're talking of this:
You have some kind of pendulum attached to a point R. In fact, the total kinetic energy is $$T=T_{P}+T_{Rot}$$ Where $T_{P}$ is the translational energy of the point P and $T_{Rot}$ the rotational energy around that point. Realize that the expression of the two energies is different, depending of your axis.
When you consider your axis system at a fixed point, you only have rotational energy, because the fixed point has not a transalation. The other common option is to take the center of mass, and, in that case, you also have to consider its translational energy. This is useful if you don't have any fixed point. However, $T$ is indepedent of the axis, so when you calculate rotational kinetic energy on P, you obtain the total energy; in the other hand, the center of mass of the system is translating and rotating (if the rod has mass), so translational energy of the ball will be lower than total energy. If the rod has no mass, then the center of mass is only translating and yes, in that case the rotational energy on P would be the same has the ball's translational energy.
To solve problems with fixed point, the best option is to take that point, because you don't have to deal with translational energy. In this particular case, P is a fixed point, so you can calculate the moment of inertia of the system on P (using Steiner's Theorem), write the position and the angular velocity as function of $\beta$, (angular velocity would be $\dot{\beta}$), and apply energy conservation. If the rod has no mass, you also can try to calculate center of mass position in terms of $\beta$, derive it to obtain the velocity and calculate the modulus. If the rod has mass, this second method becomes a difficult problem: you have to calculate center of mass of the two bodies, the inertia tensor at that point, and apply both translational and rotational energies.
I hope this will be useful. If you need some aclaration, say it.
