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Rotational Motion.

  • Thread starter Kelschul
  • Start date
  • #1
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Homework Statement


Two identical soccer balls are released from rest from the top of ramp consisting of a straight section connected to a circular section having the shape shown at right (height of ramp from where ball is released to bottom of curve is 4.0m. The radius of the curve is 1.5m). The end of the circular section of track is vertical. One ball slides down the ramp, while the other ball rolls without slipping. A soccer ball can be considered a thin-walled spherical shell.
(a)What is the speed of each ball at the bottom of the curve?
(b)What is the ratio of the normal force on the sliding ball to the normal force on the rolling ball at the bottom of the curve?
(c)What speed does each ball have when leaving the ramp?

Homework Equations


I=(2/3)MR(^2)
Wnc=((1/2)mvf(^2)+mghf)-((1/2)mv0(^2)+mgh0)



The Attempt at a Solution



Ok. I used the Wnc equation above to find the speed of the sliding ball, which I found to be 8.9m/s.
I tried to find the speed of the rolling ball by adding Iwf(^2) into the Wnc equation... but I couldn't get it to work.
w=vr, but I don't know the radius of the ball.
Do I need to find the radius, or is there another way to complete the problem?
I'm lost at what to do next... and trying to keep my eyes open :bugeye:
 

Answers and Replies

  • #2
Doc Al
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I tried to find the speed of the rolling ball by adding Iwf(^2) into the Wnc equation... but I couldn't get it to work.
w=vr, but I don't know the radius of the ball.
Do I need to find the radius, or is there another way to complete the problem?
You don't need the radius. Hint: In the rotational KE term ([itex]1/2 I \omega^2[/itex]), plug in the rotational inertia for a spherical shell (in terms of M and R) and see what happens.
 
  • #3
And v=rw, not w=vr.
 

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