Physics 8.5 Rotational Kinetic Energy (11 of 19) Ball Rolling Up an Incline (3 of 3)

[@pipertripp]

Could you think of this problem in two halves: the first half characterized by rolling w/o slipping and the second part as rolling with slipping? I am a little puzzled how energy is transferred from the rotation to potential energy in the first half of the problem, but in the second half that potential energy is transferred back, but only to the translational kinetic energy. It all makes intuitive sense, but hard to put my finger on it. Clearly in part one, the rotational velocity is a function of the translational velocity... the rolling w/o slipping. If there's no slipping, then it seems the rotational velocity must be tied directly to the translational velocity. However, in the second part of the problem, the ball is spinning freely. It's rate of rotation is NOT tied now to the translational velocity and since there are no forces acting to impede the rotation, the kinetic enery of rotation remains constant during the free fall portion of the ball's journey from start to finish. Does that seem like a reasonable explanation of what's going on with the rotational kinetic energy in this second half of the problem?

[@MichelvanBiezen]

The second part of the sentence: "and the second part as rolling with slipping" doesn't appear to describe the second part of the problem. When a car rolls up a hill, its kinetic energy is converted to potential energy and the total energy remains the same.

[@pipertripp]

What I was referring to is the fact that the ball's rotation is, in the second part of the problem, no longer related to the translational motion. In rolling without slipping, the rotational velocity is essentially a function of the translational velocity (or the other way round). If the ball is freely rotating in flight, it's rotational and translational velocities are no longer related, eh (hence why the rotation kinetic energy in part two remains contant whilst the translational velocity is actually higher than at the start of the problem)?

[@anirudhhaharsha2934]

When I calculate the rotational kinetic energy final using .5 I w^2, and using the final velocity, I get around 180J. Why is this?

[@MichelvanBiezen]

The w should be calculated from the velocity the ball had at the top of the incline.

[@anirudhhaharsha2934]

@@MichelvanBiezen I figured out the problem I used v equals rw even though that only applies in a situation where the object is rolling on the ground without slipping

[@r.m8345]

but the linear velocity changes so angular velocity has to change too deltaV=Vf-Vi deltaW=Wf-Wi deltaW=Vf/r-Vi/r and because Vf doesnt equal to Vi so has not to be 0 even when the farce doesnt make tourqe

[@MichelvanBiezen]

The video is correct.

[@r.m8345]

and when you calculated the velocity on the of the hill you didnt say that the angular velocity has to be the same so we had angular velocity even though the force of gravity in the X component doesnt make torque on the ball

[@r.m8345]

please answer my question

[@MichelvanBiezen]

At the top of the hill v = Rw. After the ball leaves the top of the hill the angular velocity doesn't change but the linear velocity does.

[@jianningluo6916]

Can I say, when the ball touch the ground or something, the angular velocity is changing with translation velocity; if the ball fall in the air, means nothing can the ball touched, so angular velocity is constant and only translation velocity is changing. Is it correct ?

[@mohammedshukri2648]

Hello, why is Vfy positive? Is it because it is in the same direction as the acceleration?

[@MichelvanBiezen]

The magnitude of any vector can only be positive. So it depends how you want to look at the velocity. As a vector is it in the negative direction. As a magnitude it is positive. Since we are calculating the kinetic energy, we use the magnitude of the velocities.

[@mohammedshukri2648]

You are the best. Thanks!

[@dancinkayley]

Heya, I was wondering why my alternative method for finding the velocity isn't viable. I understand why your method works, I just don't understand why mine doesn't. (It worked for the previous video...) W + KEo + PEo = KEf + PEf + heat loss 0 + 364J (from previous videos) + mgh = KEf + 0 + 0 KEf = 364J + (2)(9.8)(10) = 560J KEf = KEt + KEr KEf = (1/2)mv^2 + (1/2)Iw^2 560J = (1/2)(2)v^2 + (1/2)(2/5)(2)(r^2)(v^2/r^2) 560J = v^2 + (2/5)v^2 560J = v^2(1 + 2/5) v^2 = (560)/(7/5) v^2 = 400 v = 20 As you can see, I fall a bit short of the real answer. Any help would be greatly appreciated!

[@dancinkayley]

Wait now I think I get it, the rotational kinetic energy wasn't changing so I should've replaced (1/2)(2/5)(2)(r^2)(v^2/r^2) with 104J. Would my method work if the ball was rolling down a hill on the other side (as opposed to freefalling?)

[@MichelvanBiezen]

I am stuck on your first line: "from previous videos". I would like to see that line replaced with what you did here in this problem.

[@MichelvanBiezen]

I believe so, but I need to see all your work to make that determination. Did you get the correct answer?

[@dancinkayley]

The 364J was the total kinetic energy at the top of the hill. I calculated that through the following: KE (bottom of hill) = (1/2)mv^2 + (1/2)Iw^2 = (1/2)(2)(20^2) + (1/2)(2/5)(mr^2)(v^2/r^2) = 400 + (1/5)(2)(20^2) = 400 + 160 = 560J W + KE(bottom) + PE(bottom) = KE(top) + PE(top) + heat lost 0 + 560J + 0 = KE(top) + mgh + 0 KE(top) = 560J - mgh KE(top) = 560J - (2)(9.8)(10) = 560J - 196J = 364J

[@dancinkayley]

When I replaced (1/2)(2/5)(2)(r^2)(v^2/r^2) (in my original comment) with 104J, I did indeed get the correct answer.