Rotational Mechanics-Symmetric body goes up an incline.

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Homework Help Overview

The problem involves a symmetric body rolling up an inclined plane after rolling without slipping on a horizontal surface. The objective is to determine the velocity of the body as it transitions onto the incline, considering factors such as the angle of inclination and the radius of gyration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of angular impulse and the change in angular momentum, with some questioning the assumption of the normal force being mgcosθ during the collision with the incline. Others explore the implications of impulsive forces and suggest considering angular momentum about specific points to simplify the analysis.

Discussion Status

The discussion has led to various interpretations regarding the forces acting on the body and the conservation of angular momentum. Some participants have successfully navigated the problem by reconsidering the normal force and identifying strategic points for analysis, while others continue to explore the implications of their assumptions.

Contextual Notes

There is an ongoing examination of the impulsive forces at play during the transition from horizontal to inclined motion, and how these forces affect the calculations of momentum and angular momentum. Participants are also reflecting on their intuitive understanding of the problem based on experience.

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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