Sphere rolling down an incline rotational kinetic energy

AI Thread Summary
A solid steel ball with a diameter of 2.00 mm rolls down a 1.50-meter diameter loop, losing 10% of its mechanical energy. The calculations indicate that the height from which the ball must be released is approximately 0.975 meters to just make it through the loop. The discussion emphasizes the importance of considering both potential and rotational kinetic energy in the calculations. It also raises questions about the validity of the speed formula used and the necessary conditions for the ball to remain on track at the loop's top. Ultimately, the participants are trying to clarify the correct height needed for the ball to successfully navigate the loop.
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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 hight of a track before the loop needs to be 5/2 larger without a loss term and rotational kinetic energy ??
 
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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 hight of the slope if the ball did not loose any energy?

ehild
 
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