Rotational Mechanics-Symmetric body goes up an incline.

AI Thread Summary
The discussion focuses on determining the velocity of a symmetric body as it rolls up an inclined plane after rolling without slipping on a horizontal surface. Key points include the importance of correctly identifying the forces acting on the body, particularly the impulsive force from the incline, which differs from the normal force after the collision. Participants emphasize using angular impulse to analyze the problem and suggest choosing an appropriate point to simplify calculations, such as the point of contact with the incline. The final solution for the velocity, v2, is expressed in terms of the initial velocity, angle of inclination, radius, and radius of gyration. The conversation concludes with a successful resolution of the problem, highlighting the value of collaboration and problem-solving strategies.
consciousness
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Homework Statement



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

Homework Equations



I am trying to use-

Angular impulse=Change in angular momentum

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 can't calculate the external impulse
any help, ideas are much appreciated!
 
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consciousness said:
Assumed that the normal force exerted by the incline is mgcosθ. The velocity to find is v2.

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.
 
Hi Tsny! I have finally solved this question by not taking the normal force as mgcosθ. But how can we tell that it won't 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?
 
That all sounds good.

consciousness said:
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?

Yes, that's right.
 
TSny said:
That all sounds good.



Yes, that's right.

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 anyone is interested in knowing the answer it is-

v2=v1(1-(R^2(1-cosθ/R^2+K^2))
 
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