# Sphere rolling down an incline rotational kinetic energy

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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ehild
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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 ?

ehild
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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

ehild
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