Rolling Body mechanics

  • #1
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Homework Statement


In the picture ,problem 8.29

Homework Equations


Energy conservation ,velocity relation ,momentum

The Attempt at a Solution


First i defined the speed of the body relative to the ground as Vb and speed of plane relative to the ground as Vp.
From momentum ,VbcosΘ=-Vp .From the non slipping condition ,ωR-Vb=VpcosΘ.This relation seems to be wrong ,the correct aswer is (Vp+VbcosΘ)/cosΘ=ωR which I am incapable of explaining.I also included my work in the pictures
 

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Answers and Replies

  • #2
TSny
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First i defined the speed of the body relative to the ground as Vb and speed of plane relative to the ground as Vp.
From momentum ,VbcosΘ=-Vp .
Shouldn't the masses show up in the momentum equation? Also, VbcosΘ is not the horizontal component of the velocity of the ball relative to the ground since Vb does not make an angle Θ to the horizontal. The velocity of the ball relative to the ground is not in the same direction as the velocity of the ball relative to the incline. You will need to think carefully about relative velocities in this problem.

From the non slipping condition ,ωR-Vb=VpcosΘ.This relation seems to be wrong ,the correct aswer is (Vp+VbcosΘ)/cosΘ=ωR which I am incapable of explaining.I also included my work in the pictures
I don't think either of these two equations is correct.
 
  • #3
haruspex
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(Vp+VbcosΘ)/cosΘ=ωR
As TSny wrote, that is incorrect, if Vp is as you have defined it. But it does contain some useful pointers.
In the context of the question, what physical entity does ωR represent?
What about ωRcosΘ?
The signs on the velocities depend how the variables are defined. Are they positive in the same direction or in opposite directions?
Are you sure the text isn't defining Vp as a relative velocity?
 
  • #4
55
1
Yes I realized that the velocity relative to the plane is the one that points along the plane ,not the velocity relative to the ground ,and everything makes sense now.Thank you so much !
 

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