What is the Release Height for a Ball and Disk in a Loop-the-Loop Problem?

[ruffrunnr]

A small spherical ball of radius r = 1.7 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 3.1 m. The ball has mass M = 345 g.

How high above the top of the loop must it be released in order that the ball just makes it around the loop?

Repeat problem (a) for a disk. Find the ratio of the heights h for the two cases.


I used conservation of energy and set mgh = KE(rot) + KE(trans) + mg2R. Using V=sqrt(Rg), I found the minimum speed the ball must have at the top of the loop and found it to be 5.512 m/s. For the right side of the equation, I used KE(rot) = 1/2Iw(omega)^2. For a spherical ball, I =2/5MR^2.
Now I have
mgh = 1/2mv^2 +1/2(2/5)MR^2(w^2) + 2mgR
M's cancel
w = v/R
So I was left with
h = (1/2(v^2)+1/5(R^2)(v/R)^2 + 2gR)
----------------------------------
g
h = 8.37m
h-2R = H = 2.17m

That is the correct answer according the homework but then it asks for a disk with the same radius and here is where I don't get the right answer.

mgh = 1/2mv^2 +1/2Iw^2 + mg2R
mgh = 1/2mv^2 +1/2*1/2mR^2w^2 + mg2R
m's cancel

h = 1/2v^2 +1/4R^2(v/R)^2 +2gR
-----------------------------
g

Help anyone?

 

[Nate810]

For the disk, you can use the same equations as for the ball. The only difference is that for the disk, I = 1/2MR^2. So the equation becomes:h = (1/2(v^2)+1/4(R^2)(v/R)^2 + 2gR) ---------------------------------- gh = 7.25m h-2R = H = 1.05mThe ratio of heights h for the two cases is 8.37/7.25 = 1.15

 

[m2003]

I would first like to commend you on your use of conservation of energy to solve this problem. It is a great approach to solving problems involving motion and is a fundamental principle in physics.

Now, to address your question about the disk case, there are a few things to note. First, the moment of inertia for a disk is different from that of a sphere. For a disk, the moment of inertia is 1/2MR^2, which is why you see the difference in your equations.

Also, in your calculation for the disk, you have used the moment of inertia for a sphere instead of a disk. This is why you are not getting the correct answer. The correct equation should be:

mgh = 1/2mv^2 + 1/2 * 1/2MR^2 * (v/R)^2 + mg2R

Now, solving for h, we get:

h = 1/2v^2 + 1/4R^2(v/R)^2 + 2gR

Simplifying this further, we get:

h = 7/4v^2 + 2gR

Using the same approach as before, we can calculate the minimum speed required for the disk to make it around the loop:

v = sqrt(Rg) = sqrt(3.1 * 9.8) = 5.501 m/s

Substituting this value into the equation for h, we get:

h = 7/4(5.501)^2 + 2(9.8)(3.1) = 8.348 m

Therefore, the height at which the disk must be released in order to just make it around the loop is 8.348 m, which is slightly lower than the height for the spherical ball.

Now, to find the ratio of the heights, we simply divide the height for the disk by the height for the sphere:

h_disk / h_sphere = 8.348 / 8.37 = 0.997

This means that the height for the disk is approximately 99.7% of the height for the sphere. In other words, the disk must be released at a slightly lower height compared to the sphere in order to make it around the loop.

I hope this helps to clarify your doubts. Keep up the good work in using conservation of energy to solve problems!