Sphere rolling down an incline rotational kinetic energy

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SUMMARY

The discussion focuses on calculating the height from which a small solid steel ball must be released to successfully navigate a loop-the-loop with a diameter of 1.50 meters, accounting for a 10% loss in mechanical energy. The relevant equations include the kinetic energy formula KE = 0.5I(omega)^2 + 0.5mv^2 and the moment of inertia for a sphere, I = (2/5)MR^2. The final calculations indicate that the required height is approximately 0.975 meters, factoring in the energy loss and the need for sufficient speed at the top of the loop to maintain contact.

PREREQUISITES
  • Understanding of rotational kinetic energy and its calculations
  • Familiarity with the concept of mechanical energy conservation
  • Knowledge of the moment of inertia for solid spheres
  • Basic principles of circular motion and centripetal force
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  • Study the derivation of the kinetic energy formula for rolling objects
  • Learn about energy loss in mechanical systems and its implications
  • Explore the dynamics of circular motion, specifically centripetal acceleration
  • Investigate the effects of different energy loss percentages on required heights in similar problems
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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 ??
 
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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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