Velocity of Sphere After Collision with Vertical Wall

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Homework Help Overview

The discussion revolves around a sphere that rolls on a rough horizontal surface and collides elastically with a smooth vertical wall. Participants are exploring the conditions under which the sphere transitions to pure rolling backward after the collision, considering factors such as energy conservation and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the implications of an elastic collision and the role of friction in determining the sphere's velocity after the collision. There is also discussion about the meaning of "pure rolling backwards" and how rotational motion affects the sphere's behavior post-collision.

Discussion Status

The conversation is ongoing, with participants offering insights into the mechanics of the collision and the effects of friction. Some are attempting to derive relationships between variables, while others are clarifying terminology and concepts. There is no explicit consensus on the approach to take, but various lines of reasoning are being explored.

Contextual Notes

Participants note the absence of specific values for the coefficient of friction and the time period for the transition to pure rolling, which may influence their analyses. The discussion reflects a mix of assumptions and interpretations regarding the mechanics involved.

vaishakh
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A sphere rolls on a rough horizontal surface with a velocity v collides with a smooth vertical wall elastically. Find its velocity when its starts pure rolling backward.
I feel that since collision is elastic the energy must be conserved instantaneously. But however there is friction. However in how much time will the pure rolling start in backward direction is not clear. I don't understand how to start. Can anyone here please help?
 
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I've just started a topic on collisions myself today so I would be interested to see how this one pans out...do I assume collides elastically means that the coefficient of restitution = 1?
as for the velocity v, the best I could hope to do with the information given is to spend a while scratching my head and trying to devise some way of expressing v symbolically in terms of its initial speed and work done by friction...and somehow include e if it isn't 1

Wish I could be of more help.

btw...what does 'pure rolling backwards' mean? :redface:
 
Last edited:
'pure rolling backwards' means the ball is really rolling backward, not sliding.
 
Harmony said:
'pure rolling backwards' means the ball is really rolling backward, not sliding.

I'm going to try to expand on what Harmony said, but I have a question about it as well.

The ball is rolling and rolls into a wall. The ball will bounce off the wall, but remember that it has some rotational motion as well. I suspect what will happen is the ball's CM will reverse its direction but the ball will still be rotating in it's original direction. That is the rotation will be in the wrong sense, so the ball's surface will be slipping on the floor for a short time until friction can correct for it. I have not done the analysis myself to see if this is true because the time period for which this occurs will depend on the value of the coefficient of friction, which is not given.

My thought is that you can come up with an effective coefficient of friction by looking at the rolling without slipping condition. I personally have never tried this approach before, so I don't know how it will work out.

-Dan
 
Will you say I am correct?
Let the force of friction be F. Let a = F/m that is mass of object.
The translational velocity after collision would be v.
But in time t it would be v - at.
The Force F has a torque T = RF.
Therefore after time t, angular acceleration b = RF/I = RF/0.4MR^2 = 5a/2R.
Therefore w = -v/R + bt = 5at/2R - v/R.
The object starts rolling when v - at = wr = 5at/2 - v
Thus 2v = 7at/2.
thus v = 7at/4.
What is the defect here?
 
No defect that I can see offhand.
But your v is, as you said, the speed just after the collision.
to find the EVENTUAL speed, solve for at = 4v/7,
and replace the "unknown" a*t in your line 4 to find v_f = v - at ...
 

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