# Rotational and Translational Kinetic Energy - marble and bowl problem.

1. Nov 18, 2012

### binbagsss

A uniform marble rolls down a symmetric bowl, starting from rest at the top of the left side. The top of each side is a distance above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil.
a) How high would the marble go if both sides were as rough as the left side?
b) How far up the smooth side will the marble go, measured vertically from the bottom?

My attempt - I am able to attain the correct answers but concept wise could do with some help ...

a) via consevation of energy I conclude h.
However I do not really understand why energy is conserved in such cases 'roll without slipping'.

( I have seen a basic derivation which considers the two components of the velocity of the marble - the rotational and the translational, and setting this sum to zero to deduce that v=wr[1] )

However I do not understand why, in the conservation of energy equation , we do not include the work done via friction due to the rough surface.

b) I first of all calculate the speed at the bottom via consevation of energy - equating the loss of GPE = KE ROTATIONAL + KE TRANSLATIONAL, and using the above deduction [1] which follows from the nature of the surface and marble's mutual. behaviour.

Then using this speed, equated the KE at the bottom = GPE GAIN, knowing KE at max height = 0.

However again, I do not really understand the principles behind this- why is rotational kinetic energy instantly eliminated as soon as the body slips - How does slipping - no frition - cause motion to be purely translational .? (That's the cnclusion I reach as this led me to the right answer).

Anyone who could shed some light on this, really aprreciated, thanks alot !

2. Nov 18, 2012

### TSny

1. To understand why the friction force doesn't do any work, consider what rolling without slipping means in terms of the instantaneous velocity of the point of the marble that is instantaneously in contact with the surface of the bowl.

2. When the marble reaches the oiled surface and begins to slip, the motion does not become purely translational. When the friction force goes to zero, there is no longer any torque on the marble about the center of the marble. So, what does that imply about the behavior of the angular velocity of the marble after it begins to slip?

Last edited: Nov 18, 2012
3. Nov 18, 2012

### binbagsss

Thanks very much for the feedback. So
1) Because the friction force is perpendicular to the marble's motion at any instant.
2) The marble has zero net torque and so ω will remain unchanged?

4. Nov 18, 2012

### TSny

The point of the marble that is instantaneously in contact with the surface has zero velocity for rolling without slipping. This means that the friction force never "acts through a distance", so it doesn't do any work on the marble. You can also think of it in terms of the formula P = F*v for the power produced by a force. No power if v = 0, so no energy transferred to the marble by the friction force.
Yes. Good.

5. Nov 18, 2012

### rcgldr

Static friction refers to a Newton third law pair of forces, the somewhat upwards force from the rough surface of the bowl exerted onto the surface of the marble, and the equal but opposing somewhat downwards reaction force exerted by the surface of the marble onto the rough surface of the bowl. Static friction doesn't perform any work, but the integral of the somewhat upwards force (as a function of distance) from the rough surface of the bowl across the distance rolled equals the gain in angular kinetic energy (and also m g h - gain in linear kinetic energy). At the bottom of the bowl the total kinetic energy (linear and angular) gained is equal to m g h. If the right half of the bowl is also rough, and there are no losses, then all of the total energy is converted back into gravitational potential energy m g h, and you don't need to calculate how much of the total energy was angular or linear at the bottom of the bowl to solve part a).

It is not elimnated, but angular kinetic energy will remain constant (the ball continues to rotate at constant angular velocity) once the ball reaches the zero friction part of the bowl, so only the linear kinetic energy will be translated back into gravitational potential energy m g h, while the angular kinetic energy remains constant.

Last edited: Nov 18, 2012