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Energy, Angular Momentum, Torque, solid ball rolling down loop track? help? 
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#1
Nov1612, 02:41 AM

P: 172

A solid brass ball of mass .280g will roll smoothly along a looptheloop track when released from rest along the straight section. The circular loop has radius R = 14.0 cm, and the ball has radius r<<R.
(a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop? If the ball is released at height h = 6.00R, what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball a point Q? Solution and with diagram on first page, #11.8 www.monmsci.net/~fasano/phys1/Chapter_11_09.pdf first what does r<<R mean?? is that the same as r<R? 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? Also where did 2R come from, the problem never mention anything about 2R. 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? thanks! 


#2
Nov1612, 03:13 AM

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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 hoolahoop sized loop rather than a golf ball on a golfhole sized loop. (technically a plug is smaller than a plughole right?) BTW: they are actually telling you that h=7R: compare the last two relations again  the RHS has also changed. 


#3
Nov1712, 01:02 AM

P: 172

sorry im having a lot trouble with this problem :( 


#4
Nov1712, 09:22 PM

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Energy, Angular Momentum, Torque, solid ball rolling down loop track? help?
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 airresistance 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 rampheight is equal to the height of the loop and we can write h=2R. What would happen in that case? 


#5
Nov1712, 11:49 PM

P: 172

i think i get it. i attached a picture.



#6
Nov1812, 01:13 AM

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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. 


#7
Nov1812, 11:06 PM

P: 172

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) 


#8
Nov1812, 11:08 PM

P: 172

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) 


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