How High Must the Brass Ball Start to Complete the Loop-the-Loop?

AI Thread Summary
The discussion focuses on calculating the height from which a brass ball must be released to complete a loop-the-loop track. The user initially calculates the necessary height but finds their answer slightly off from the book's solution. They consider whether to include rotational inertia in their kinetic energy calculations, which leads to confusion about the ball's radius. A participant clarifies that the radius can be treated as an unknown that will cancel out in the equations, allowing the user to relate angular velocity to linear speed. Ultimately, the user successfully resolves their calculations after this guidance.
labgoggles
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Hi everyone, I've been working on this problem for a while now, and I was hoping someone here could point me in the right direction. Here goes:

1. Homework Statement
A solid brass ball of mass 0.280 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R= 14.0, and the ball has radius r <<R. (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop?

I don't have units of R, but I think it's in cm.

Homework Equations


At the top of the loop, FN + mg = ma
Fc = mv2 / R where R is radius of the track

Ui + Ki = Uf + Kf

The Attempt at a Solution


First, using FN + mg = ma, I know ma is (mv2)/R, and (mv2)/R = mg if the ball is going to fall at the top, from (mv2)/R - mg = FN.

I plug in R = .14m, m= .280 X 10-3g, and g to get v2 = 1.37 m/s. I think it's correct up to here...

However, in Ui + Ki = Uf + Kf, initial U = mgh, final U =mg(2R) because the ball is at the top of the loop, initial K = 0, but final K = 1/2mv2. When I plug everything in and m cancels out,

gh = 1/2v2 + g(2R)

(9.8m/s2) = 1/2(1.37m2/s2) + (9.8m/s2)(2 X .14 m)

and h = .350 m or 35.0 cm. My answer is close to the book's, which gives h = 37.8 cm, but I think I am missing something. Do I need to include rotational inertia, (1/2)Iω2 as part of K final? If I did, it looks like I would need to find I and omega, and I need the radius of the ball to do that (not given). I only have its mass and calculated velocity. Please let me know if I need to give more information and clarify, thank you!
 
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labgoggles said:
Do I need to include rotational inertia, (1/2)Iω2 as part of K final? If I did, it looks like I would need to find I and omega, and I need the radius of the ball to do that (not given). I only have its mass and calculated velocity.
Yes, that's the error.
You don't need to know r. Put it in as an unknown and it will cancel out. For a rolling uniform ball, there is a constant fraction of its KE that is in the rotation. Relate omega to linear speed in the usual way.
 
Thank you so much! I was able to work it out.
 

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