- #1
labgoggles
- 4
- 0
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.
At the top of the loop, FN + mg = ma
Fc = mv2 / R where R is radius of the track
Ui + Ki = Uf + Kf
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!
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!