Solving Rolling and Collision Homework: Find Veloc. & Ang. Vel. of Sphere

[ritwik06]

Homework Statement

In the given arrangement a sphere of mass "m" rolling on a rough surface strikes a rough inclined plane. Find the velocity of centre of mas and the angular velocity of sphere if just after collision, it starts ascending up the incline plane and performs rolling.

http://img228.imageshack.us/img228/7488/picw.jpg

The Attempt at a Solution

When the sphere is just about to ascend the incline plane, it will be in contact with two surfaces and hence two contact forces will act on it just at the moment when it is about to ascend. But both of these forces will pass through the centre of mass, so their net torque about centre of mass will be zero. The angular velocity omega will remain same.

To find the velocity of the sphere I apply conservation of momentum about point P (with which the sphere will be in contact at the time of ascending).
Initial angular momentum (mVoRcos (theta)+ I*omega)=Final momentum(mV'R+I*omega)
V'=Vo cos (theta)

Where is the flaw in my solutions?

 

[tiny-tim]

Hi Ritwik! Thanks for the PM.

To find the velocity of the sphere I apply conservation of momentum about point P (with which the sphere will be in contact at the time of ascending).
Initial angular momentum (mVoRcos (theta)+ I*omega)=Final momentum(mV'R+I*omega)
V'=Vo cos (theta)

Where is the flaw in my solutions?

Do you keep missing out the word "angular" before "momentum" accidentally or deliberately?

Anyway, you seem to be adding ordinary (linear) momentum to angular momentum in the same equation …

you can't do that!

 

[ritwik06]

Hi Ritwik! Thanks for the PM. Do you keep missing out the word "angular" before "momentum" accidentally or deliberately?

It was accidental. I meant angular momentum. I am sorry!

Anyway, you seem to be adding ordinary (linear) momentum to angular momentum in the same equation …

you can't do that!

where?
$$m*V_{o}*R* cos \theta+ I*\omega=m*V'*R+I*\omega$$
So do you mean conservation of angular momentum cannot be applied here?

I would be glad if u could tell me which steps were wrong.
1.The angular velocity omega will remain same. Is this wrong?
2. Angular momentum will be conserved about point P. Is this one wrong?

thanks a lot,
regars,
Ritwik

 

[tiny-tim]

It was accidental. I meant angular momentum. I am sorry!


where?
$$m*V_{o}*R* cos \theta+ I*\omega=m*V'*R+I*\omega$$
So do you mean conservation of angular momentum cannot be applied here?

I would be glad if u could tell me which steps were wrong.
1.The angular velocity omega will remain same. Is this wrong?
2. Angular momentum will be conserved about point P. Is this one wrong?

thanks a lot,
regars,
Ritwik

Sorry, I got confused

The extra term looked wrong, and I couldn't see why …

On looking at it again, you've used the correct Iω angular momentum, but then added to it an mvr as if, in addition to the sphere rolling, there was an equal mass concentrated at the centre of the sphere …

so to find the angular momentum about the contact point, you've added the c.o.m. angular momentum to the moment of the linear momentum about the contact point.

This does work (because of the parallel axis theorem), but only because the sphere is rolling … in any other case, you wouldn't be able to do that.

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The question itself is not of a type I've seen before

You can easily get the impulsive torque by taking moments about the centre of the sphere, and also impulsive linear force by taking momentum components in two perpendicular directions … but that doesn't seem enough to eliminate all the unknowns so as to give v_f as a function of v_i

There must be a reason why we can ignore one of the impulses … perhaps the friction impulse with the ground … which I can't see