Sphere rolling down an incline rotational kinetic energy

In summary: The ball would need to be released higher than 5/2 the height of the track to just make it through the top of the loop.
  • #1
rollorcoaster
3
0
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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  • #2
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
 
  • #3
oh ok
Fc = mg

v^2 = rg ??

then i get 1.97 meters is that right ?
 
  • #4
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
 
  • #5
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
 
  • #6
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
 

What is rotational kinetic energy?

Rotational kinetic energy is the energy an object possesses due to its rotation around an axis. It is calculated as 1/2 * I * ω², where I is the moment of inertia and ω is the angular velocity.

How is rotational kinetic energy related to a sphere rolling down an incline?

When a sphere is rolling down an incline, it is both rotating and translating. As it rotates, it gains rotational kinetic energy, which is then converted into translational kinetic energy as it moves down the incline.

What factors affect the rotational kinetic energy of a sphere rolling down an incline?

The rotational kinetic energy of a sphere rolling down an incline is affected by the mass, radius, and moment of inertia of the sphere, as well as the angle of the incline and the friction present.

How is the moment of inertia of a sphere calculated?

The moment of inertia of a sphere is calculated as 2/5 * m * r², where m is the mass of the sphere and r is the radius. This formula assumes that the mass is evenly distributed throughout the sphere.

Can the rotational kinetic energy of a sphere rolling down an incline be greater than its translational kinetic energy?

No, the rotational kinetic energy of a sphere rolling down an incline can never be greater than its translational kinetic energy. This is because some of the rotational kinetic energy is always converted into translational kinetic energy due to friction and rolling resistance.

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