# Energy, Angular Momentum, Torque, solid ball rolling down loop track? help?

 P: 172 A solid brass ball of mass .280g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 14.0 cm, and the ball has radius r<
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 Quote by nchin A solid brass ball of mass .280g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 14.0 cm, and the ball has radius r<
Yes - but more so.
It means that the radius of the ball is very much smaller than the radius of the loop - so you are thinking marble on a hoola-hoop sized loop rather than a golf ball on a golf-hole sized loop. (technically a plug is smaller than a plughole right?)
 a) so the initial E is mgh when the ball is at the very top. Ei = Ef mgh = 1/2mv^2 + 1/2Iω^2 + mg(2R) So what is mg(2R)?? how can we substitute 2R for h?
The ball starts out with an initial height h - then it goes down to the ground, then it goes up to the top of the loop .... how high is the top of the loop from the ground?
 Also where did 2R come from, the problem never mention anything about 2R.
But they did mention an R. What is R?
 b) mg(6R) = mg (R) + 1/2mv^2 + 1/2Iω^2 so i understand that the problem states that h = 6R on the left hand side, but why is the right hand side only R (mgR). What does that R mean?
You have to read the question - they tell you right at the start.
BTW: they are actually telling you that h=7R: compare the last two relations again - the RHS has also changed.
P: 172
 Quote by Simon Bridge The ball starts out with an initial height h - then it goes down to the ground, then it goes up to the top of the loop .... how high is the top of the loop from the ground?
i still dont understand how do we figure out how high is the top of the loop from the ground? The problem mentions "The circular loop track has radius R = 14.0cm" so how do you get 2R? 2R would be 28cm?

 Quote by Simon Bridge But they did mention an R. What is R?You have to read the question - they tell you right at the start.
R is the radius of circular loop track. But the problem says R = 14.0cm. Why is it 2R?

 Quote by Simon Bridge BTW: they are actually telling you that h=7R: compare the last two relations again - the RHS has also changed.
how is it 7R?

sorry im having a lot trouble with this problem :(

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Energy, Angular Momentum, Torque, solid ball rolling down loop track? help?

 Quote by nchin i still dont understand how do we figure out how high is the top of the loop from the ground? The problem mentions "The circular loop track has radius R = 14.0cm" so how do you get 2R? 2R would be 28cm?
Well done.
 R is the radius of circular loop track. But the problem says R = 14.0cm. Why is it 2R?
The loop sits on the ground. R is the distance from the ground to the center ... the ball does not just go from the ground to the center does it? How high is the highest point on the track?
 how is it 7R?
Like I said - compare with the previous equation.

To understand this problem you have to remember that gravity does not care which path the ball takes as it goes up and down. All gravity cares about are the vertical distances. If the ball ran down one ramp and up another, it would end up at the same height it started (ignoring air-resistance etc). It does not matter what angle the tracks are.

In this problem, the ball starts out at some initial height y=y0=h, then it gains kinetic energy by falling (along a track) to the ground (at y=0). The energy it gains doing this is mgh. This is how much energy is available for the rest of the motion.

The next thing that happens is it enters a loop with a radius of R.
The loop sits of the floor, so the very top of the loop is a distance y=y1=2R high.

This is where you get stuck - so I'll stop there for now. Sketch out the apparatus: draw in a ramp that feeds into a loop and draw in the different y positions of everything. Then you'll see it.

if y0=y1 then the ramp-height is equal to the height of the loop and we can write h=2R.
What would happen in that case?
 P: 172 i think i get it. i attached a picture. Attached Thumbnails
 Homework Sci Advisor HW Helper Thanks P: 12,962 Good - you've got where the 2R comes in :) Drawing stuff is a good idea in physics - even if you are not asked to. At this stage - leave the ramp height as h ... this is what you are trying to find right? Still aiming to understand the problem, lets explore a bit: If h=2R - so the ramp is as high as the top of the loop, what do you think will happen? At the bottom of the ramp, the energy is entirely kinetic. When the ball goes around the loop - it has circular motion: you will have studied circular motion recently. At the top of the loop, there is a minimum force that has to be in place to keep the ball moving in the same circle.
P: 172
 Quote by Simon Bridge Good - you've got where the 2R comes in :) Drawing stuff is a good idea in physics - even if you are not asked to. At this stage - leave the ramp height as h ... this is what you are trying to find right? Still aiming to understand the problem, lets explore a bit: If h=2R - so the ramp is as high as the top of the loop, what do you think will happen? At the bottom of the ramp, the energy is entirely kinetic. When the ball goes around the loop - it has circular motion: you will have studied circular motion recently. At the top of the loop, there is a minimum force that has to be in place to keep the ball moving in the same circle.
I think I understand this problem now :)

So, energy is conserved

To find h,

Total energy = KE translational + KE rotational + PE on top of circle

mgh = 1/2 mv^2 + 1/2Iω^2 + mg(2R)
P: 172
 Quote by Simon Bridge Good - you've got where the 2R comes in :) Drawing stuff is a good idea in physics - even if you are not asked to. At this stage - leave the ramp height as h ... this is what you are trying to find right? Still aiming to understand the problem, lets explore a bit: If h=2R - so the ramp is as high as the top of the loop, what do you think will happen? At the bottom of the ramp, the energy is entirely kinetic. When the ball goes around the loop - it has circular motion: you will have studied circular motion recently. At the top of the loop, there is a minimum force that has to be in place to keep the ball moving in the same circle.
I think I understand this problem now :)

So, energy is conserved

To find h,

Total energy = KE translational + KE rotational + PE on top of circle

mgh = 1/2 mv^2 + 1/2Iω^2 + mg(2R)
 Homework Sci Advisor HW Helper Thanks P: 12,962 Not bad ... the rotational and tangential speed are related - since the ball is rolling without slipping. But you need to find what that speed has to be.

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