# Homework Help: Rotational Mechanics-Symmetric body goes up an incline.

1. Apr 19, 2013

### consciousness

1. The problem statement, all variables and given/known data

Find the velocity with which a symmetric body of radius "R" starts to roll up an inclined plane if it was rolling without slipping on a horizontal plane with velocity "v1" and reached the incline.
Angle of inclination of plane with horizontal=θ
Radius of gyration of the body=k

2. Relevant equations

I am trying to use-

Angular impulse=Change in angular momentum

3. The attempt at a solution

Assumed that the normal force exerted by the incline is mgcosθ. The velocity to find is v2.
I am taking Angular momentum about the center of mass but I cant calculate the external impulse
any help, ideas are much appreciated!

2. Apr 19, 2013

### TSny

When the object collides with the inclined plane there will be an impulsive force from the inclined plane. The normal component of this force will not be mgcosθ. Only after the collision is over will the normal component of force be mgcosθ. The impulsive force will also have a component parallel to the inclined plane.

If you apply "angular impulse = change in angular momentum" about a judiciously chosen point, then you can avoid having to deal with the unknown impulsive force.

3. Apr 20, 2013

### consciousness

Hi Tsny! I have finally solved this question by not taking the normal force as mgcosθ. But how can we tell that it wont be so? Intuition ? Experience?

I took the impulse to be general and took two components one parallel and one normal to the incline.

Since i was taking the Ang. momentum about the center the component of linear impulse normal to the plane will have no effect as it passes through the center. We can calculate the parallel component by the relation-

Linear impulse along incline=change in momentum along incline= m(v2-v1cosθ)

And then-

Linear impulse*Perpendicular distance=Angular Impulse

simplify to get the answer in terms of θ,k,R.

But i would prefer if

"about a judiciously chosen point, then you can avoid having to deal with the unknown impulsive force."

I can think of one such point it is the point where the body first touches the incline am I right?

4. Apr 20, 2013

### TSny

That all sounds good.

Yes, that's right.

5. Apr 20, 2013

### consciousness

Apart from the impulsive forces by the incline there are no other such forces on the body. So the angular momentum about this point remains conserved.

Applying conservation of angular momentum and some geometry involving tangents to a circle I got the same answer. This time in a very short time! I was initially quite apprehensive about this question. Thanks a lot TSny!

If any one is interested in knowing the answer it is-

v2=v1(1-(R^2(1-cosθ/R^2+K^2))