Rolling Without Slipping/Loop Problem

  • Thread starter mburt3
  • Start date
  • #1
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Homework Statement


A small, solid sphere of mass 0.6kg and radius 27 cm rolls without slipping along the track consisting of slope (at an angle of 60degrees from horizontal) and loop-the-loop with radius 2.65m at the end of the slope. It starts from rest near the top of the track at a height, h, where h is large compared to 27 cm. What is the minimum value of h such that the sphere completes the loop?


Homework Equations


mgh=1/2mv^2 + 1/2Iw^2
mv^2/r=mg
1/2mv(top)^2 + mgr2 = 1/2mv(bottom)^2 + 0


The Attempt at a Solution


I solved for v at the top of the loop to be v=square root of (gR)
v(top)=5.096m/s

Then I used the equation 1/2mv(top)^2 + mgr2 = 1/2mv(bottom)^2 + 0
and solved for v at the bottom.
v(bottom)= 11.4 m/s
Then I plugged this into the equation: mgh=1/2mv(bottom)^2 + 1/2Iw^2
and simplified it to be
gh= v(bottom)^2 + 2/5v(bottom)^2
and solved for h to get 9.27 m.

Where did I mess up?
 

Answers and Replies

  • #2
166
0
You have to include rotational kinetic energy in all your energy conservation calculations.
 
  • #3
23
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can i still solve for the velocity at the top in the same way? If not, how would I incorporate rotational velocity into that?
 
  • #4
344
1
Draw a FBD.

1) mg-Fn=ma=mv^2/R
2) mgh=mg(2R)+1/2mv^2+1/2Iw^2

then solve for h

I got h=27/10R
 
  • #5
166
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You haven't explained what that equation is.

If you obtained it by considering the motion of the center of mass alone, there is no need to include the rotational velocity. Remember, that is a force balance equation, not an energy conservation relation.
 
  • #6
23
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Yea I got 7.15 which is 27/10R, but I got it wrong on my online hw. Isn't it true that the force of normal reaction at the top of the loop is zero if it is just going fast enough to make it over. If so, wouldn't mg=mv^2/R.
 
  • #7
344
1
You haven't explained what that equation is.

If you obtained it by considering the motion of the center of mass alone, there is no need to include the rotational velocity. Remember, that is a force balance equation, not an energy conservation relation.

huh? Don't you have to include rotational energy?


Yea I got 7.15 which is 27/10R, but I got it wrong on my online hw. Isn't it true that the force of normal reaction at the top of the loop is zero if it is just going fast enough to make it over. If so, wouldn't mg=mv^2/R.

I assumed that the Fn is zero, can't really expain why though. I'm sure that assuming it is zero correct, I did my problem from a text book and the answer is the same as the text book. (my question is just in general form)
 
  • #8
166
0
huh? Don't you have to include rotational energy?

My previous post just referred to the part about finding the velocity at the top. "That equation" referred to the force-balance equation that you wrote down.

Your solutions look fine to me. The only problem might be that the radius of the sphere is not really negligible in comparison to the radius of the loop. You might be expected to take that into account (?)
 
  • #9
23
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ok thanks for all of your help!!!!
 

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