Circular rotational KE motionl.

[tubworld]

A tennis ball is set rolling without slipping on a horizontal section of a track. It rolls into a vertical loop and goes up the loop.

I was just wondering: Suppose that the static friction betweeen the ball and the track were negligible, so that the ball slid instead of rolling, would its speed be higher or lower or the same as that of the loop?

For me, my explanation: rotational KE = 0 (because the ball isn't rolling, but slipping) and linear KE is the same, the initial total energy would be lesser than the original question, the speed of the ball at the top should be lesser then.

But from another pt of argument, if static friction is negligible, it means that the pt of the ball which is in contact with the ground at any pt is 0. As such, rotational KE remains the same at the bottom of the track and at the top of the loop. Rotational KE (initial) = rotational KE (Final). Thus we need not consider the rotational KE in the Conservation of energy in the loop. Thus, we only need to consider translational KE and the potential energy at the start and at the top, ignoring rotational KE.

Is this the right way? Or is there another explanation for the speed at the top of the loop?

 

[anathema]

Hello, thank you for your post and interesting question. I would like to provide some insights and clarification on the scenario you have described.

Firstly, you are correct in your explanation that if the static friction between the ball and the track were negligible, the ball would slide instead of roll. This would result in a decrease in its initial total energy, as the rotational kinetic energy would be zero. Therefore, the speed of the ball at the top of the loop would be lower than if it were rolling without slipping.

However, there is another factor to consider in this scenario. When the ball is rolling without slipping, it experiences both translational and rotational motion, which contribute to its kinetic energy. But when the ball is sliding, it only has translational motion and therefore, its kinetic energy is solely determined by its linear speed.

Now, in the case of a vertical loop, as the ball reaches the top of the loop, it has to overcome the force of gravity pulling it down. This means that its kinetic energy will be converted into potential energy. In the case of the ball sliding, its initial kinetic energy is lower compared to when it is rolling without slipping. Therefore, it will have less energy to overcome the force of gravity and will reach the top of the loop at a slower speed.

To summarize, in the scenario where the static friction is negligible and the ball slides instead of rolling, its initial kinetic energy and speed will be lower. This will result in a slower speed at the top of the loop, as it has less energy to overcome the force of gravity. I hope this explanation helps to clarify your understanding. If you have any further questions, please do not hesitate to ask.

 

[quicknote]

Your explanation is correct. When the ball is rolling without slipping, it has both translational and rotational kinetic energy. However, when it is sliding without friction, it only has translational kinetic energy. This means that the total initial energy of the ball is lower in the case of sliding, as there is no rotational kinetic energy. Therefore, the ball would have a lower speed at the top of the loop compared to when it is rolling without slipping.

Another way to think about it is that when the ball is rolling without slipping, it has a combination of linear and angular velocity. This allows it to maintain its speed as it goes up the loop, since the angular velocity helps to counteract the decrease in linear velocity due to gravity. However, when the ball is sliding without friction, it only has linear velocity. This means that it would not be able to maintain its speed as it goes up the loop and would slow down due to gravity.

In conclusion, the speed of the ball at the top of the loop would be lower if it is sliding without friction compared to when it is rolling without slipping. This is because the ball has less initial energy and is unable to maintain its speed due to the absence of rotational kinetic energy.