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

  • Thread starter Thread starter ritwik06
  • Start date Start date
  • Tags Tags
    Collision Rolling
AI Thread Summary
The discussion focuses on solving a physics problem involving a sphere rolling on a rough surface and colliding with an inclined plane. Key points include the conservation of angular momentum and the confusion around mixing linear and angular momentum in calculations. Participants clarify that while angular momentum can be conserved about the contact point, the approach taken in the calculations may lead to errors. The challenge lies in determining the impulsive forces and torques acting on the sphere during the collision. Overall, the problem requires careful consideration of the dynamics involved in rolling motion and collision mechanics.
ritwik06
Messages
577
Reaction score
0

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?
 
Last edited by a moderator:
Physics news on Phys.org
Hi Ritwik! Thanks for the PM. :smile:
ritwik06 said:
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? :confused:

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

you can't do that! :wink:
 
tiny-tim said:
Hi Ritwik! Thanks for the PM. :smile:Do you keep missing out the word "angular" before "momentum" accidentally or deliberately? :confused:

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! :wink:

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
 
ritwik06 said:
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 :redface:

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. :wink:

-------​

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

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 vf as a function of vi :frown:

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 :frown:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top