Sphere rolling down an incline rotational kinetic energy

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Homework Help Overview

The discussion revolves around a solid steel ball rolling down an incline and into a loop-the-loop, focusing on the conversion of mechanical energy and the role of rotational kinetic energy. Participants are examining the conditions required for the ball to successfully navigate the loop, considering energy losses and the necessary height for release.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, questioning the validity of specific formulas used to determine the ball's speed and height. There are inquiries about the necessary conditions for the ball to stay on track at the top of the loop and the implications of energy loss.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem and questioning the calculations presented. Some guidance has been offered regarding the initial potential energy and the effects of energy loss, but there is no consensus on the correct height for the ball's release.

Contextual Notes

Participants note the 10% energy loss and the need to consider both potential and kinetic energy in their calculations. There is also mention of ambiguity in the problem statement regarding the reference point for potential energy.

rollorcoaster
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A small diameter (2.00 mm), solid steel ball rolls from rest, without slipping, down a track and into a loop-the-loop of 1.50 meters diameter. Between the starting point on the track and the top of the loop the ball converts 10.0% of its initial mechanical energy into other forms of energy. From how high above the ground must the ball be released in order to just make it through the top of the loop? Consider the diameter of the ball to be small compared to the diameter of the loop but don't forget to consider rotational kinetic energy! KE = .5I(omega)^2 + .5mv^2

i of sphere 2/5MR^2
d = 1.5m

mgh - mgd = [1/5mR^2 (V/R)^2 + .5mV^2](.9)

v = [2g(h - d)]^(1/2)
gh - gd = [2/5g(h - d)+ g(h - d)](.9)
h - d = 7/5(h - d)(.9)
h = [(-7/5)(.9)d + d]/(1-(7/5)(.9))
h = .975

when the height of a track before the loop needs to be 5/2 larger without a loss term and rotational kinetic energy ??
 
Last edited:
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Why do you use the formula v = [2g(h - d)]^(1/2)'? When is it valid?

What should be the speed at the top of the loop so as the ball stay on track? ehild
 
oh ok
Fc = mg

v^2 = rg ??

then i get 1.97 meters is that right ?
 
Take care where you put that 0.9. The ball loses 10 % of its initial mechanical energy, what is the mechanical energy initially? Yes, you can count the potential energy either from the top or the bottom of the loop, (the writer of the problem could have been a bit more specific) but the ball moves down at the bottom of the track, so I would count the initial potential energy from the bottom of the loop: PE(initial) = mgh. Anyway, initially the ball has only potential energy, and 90% is converted to KE +PE at the top of the loop.


ehild
 
but is what i have the right answer or did i make a mistake somewhere
??

it doesn't seem like that is high enough for the ball to make it thru the loop
 
No, I do not think that your answer is right.
What would be the necessary height of the slope if the ball did not loose any energy?

ehild
 

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