# Loop the Loop

## Homework Statement

A small ball of radius 1.9 cm rolls without slipping down into a loop-the-loop of radius 3.7 m. The ball has mass of 353 gm.

How high above the top of the loop must it be released in order that the ball just makes it around the loop, and what is the total kinetic energy (translational plus rotational) at the highest point of the loop?

## Homework Equations

K = 1/2 mv^2 + I W^2
K = 1/2 mv^2 (1 + I/mr^2)

## The Attempt at a Solution

Honestly, I have no idea how to attack this problem. I know that energy is conserved, but really have no idea how to go about solving this problem. Can someone give me an idea of how to start going about this?

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Delphi51
Homework Helper
The basic method is to realize that, at the top, the speed must be sufficient so that the force of gravity is equal to the centripetal force. Those energy formulas should give you a relationship between the height and speed so you can find the height.

Ok, so:
Force of gravity = Centripetal Force
Centripetal acceleration = Radius times Angular Velocity squared.
Angular velocity = tangential velocity divided by radius.
So: Centripetal Acceleration = R (V/R)^2
Acp= 9.81
9.81 = 3.7 (V/3.7)^2
2.65 = V^2 / 13.69
36.297 = V^2
6.024 = V

Since Vf^2 = 2gy

36.297=2(9.81 y)
18.1485 = 9.81 Y
1.85 = Y
Where did I go wrong?

Delphi51
Homework Helper
The velocity calc looks good, though I think you should reduce the radius by .019 so that it is the radius of the circular path travelled by the center of mass of the ball.

The v^2 = 2gy is oversimplified - the rotational energy must also be taken into account.

So... one step closer, I hope.

MGH = 1/2 M V^2 (1 +(I/mr^2))
3.46 H = .1765 (36.11) (1+I/4.783)
(V and are were recalculated, per your suggestion.)
I = 7/5 MR^2
I = 7/5 .353 .019^2
I = 1.776 x 10^-4

3.46H = 6.37 (1+ 1.776 x 10^-4 / 4.783)
3.45 H = 6.37 (1.0004)
H = 1.846
No, again.
What did I do this time?

hey im working on the same problem as you
remember: v=sqrt[g(R-r)]/2 = omega
plug this into MGH = 1/2 M V^2 (1 +(I/mr^2))
You can figure out the rest.
However i am stuck on (a), what is the total kinetic energy of the ball (translational plus rotational) at the top of the loop?

if you search. this question has be answered before.

Delphi51
Homework Helper
The total kinetic (including rotational) energy is 1/2 M V^2 (1 +(I/mr^2))
where V is the speed necessary to make the centripetal force equal to the gravitational force.

I thought so. So what is wrong with:

K = 1/2 M V^2 (1 +(I/mr^2))
K = .1765 (36.11) (1+I/4.783)

I = 7/5 MR^2
I = 7/5 .353 .019^2
I = 1.776 x 10^-4

K = 6.37 (1+ 1.776 x 10^-4 / 4.783)
K = 6.37 (1.0004)
K = 6.37
?

Delphi51
Homework Helper
K = 1/2 M V^2 (1 +(I/mr^2))
K = .1765 (36.11) (1+I/4.783)
I have been making errors in calcs lately, so please check me . . . looks like that 4.783 is wrong. I get mr^2 = .353*.019^2 = .0001274 rather than 4.783.

I checked everything and I fear I made a mistake earlier in suggesting you subtract the radius of the ball from the radius of the loop. I should have said ADDED, not subtracted. Or maybe left alone.

I get E = 15.4 using radius 3.7, velocity = 6.02.
With radius 3.7-.019, I get v = 6.04 and E is still 15.4 to 3 digit accuracy.
Don't count on my calculations!

Well, the professor told us that the computer just had the wrong answer for this problem.
Whoever was inputing the problem forgot to subtract the radius of the smaller ball from the radius of the loop. So your calculations weren't the problem, it was the computer. Thanks for your help.