Movement along and off a barrel hoop

[Bonulo]

I want help for the final part of the solution to the following problem (it's initial situation being illustrated in http://chaos.fys.dtu.dk/fysik1/hjemmeopgave5.pdf - the diagram at "Opgave 2"):

"A small brick (B in the diagram) is at the bottom of the inside of half a barrel hoop that has a smooth surface. Another brick (A in the diagram) moves right on a smooth table with the speed 2*v1. The bricks A and B colide completely inelastically. The masses of the bricks are given in the diagram (both m/2)."

a) Determine the velocity of the bricks after the collision..

This is quite easy - the speed is v2 = v1/2 (the velocity's directed to the right hand side), since their masses are equal.

"For the initial velocity of brick A the inequality 2gR < v1^2 < 4gR is valid. R is the radius of the hoop."

b) Explain why the bricks can't make it to the top of the barrel hoop.

Here I've used the given inequality to compare with K < U(top) - which I have to show is true - where K is the kinetic energy of the bricks (seen as a particle), K = 1/2*m*(v1/2)^2, and U(top) is the potential energy they would have in the top if they could reach it. By showing that this potential energy will be greater than any kinetic energy the bricks can achieve, I hope to have solved the problem, but I'm not sure.

c) Determine where the bricks lose contact with the inside of the barrel hoop when v1^2 = 3gR. Write the position at an angle Theta, as defined in the diagram.

I'm stuck here. How do I determine what condition must exist for them to lose contact in the circular movement? Some has asserted that the problem can be solved by finding the maximum potential energy U, using that the bricks will drop immediatly (along the y-axis, vertically) upon losing contact. Some students assert that that can't be done, and want instead just to find the position using that n = 0, where n is the normal force from the hoop. If there is one at all.

There are about as many results, so far, as there are student groups, so we're many looking for the correct answer which would be muchly appreciated :)

 

[Doc Al]

b) Explain why the bricks can't make it to the top of the barrel hoop.
Here I've used the given inequality to compare with K < U(top) - which I have to show is true - where K is the kinetic energy of the bricks (seen as a particle), K = 1/2*m*(v1/2)^2, and U(top) is the potential energy they would have in the top if they could reach it. By showing that this potential energy will be greater than any kinetic energy the bricks can achieve, I hope to have solved the problem, but I'm not sure.

If the KE of the bricks at point B is less than mg(2R), then the bricks will certainly not make it to the top. (But note that just having KE = mg(2R) is insufficient for the bricks to maintain contact.)

c) Determine where the bricks lose contact with the inside of the barrel hoop when v1^2 = 3gR. Write the position at an angle Theta, as defined in the diagram.

I'm stuck here. How do I determine what condition must exist for them to lose contact in the circular movement? Some has asserted that the problem can be solved by finding the maximum potential energy U, using that the bricks will drop immediatly (along the y-axis, vertically) upon losing contact. Some students assert that that can't be done, and want instead just to find the position using that n = 0, where n is the normal force from the hoop. If there is one at all.

I'm not going to just give you the answer ( ) but I'll give you a big hint. As some of your fellow students have suggested, find the point where the normal force goes to zero: that's what it means to lose contact. To do that, you'll need to apply Newton's 2nd law, realizing that if the bricks are to maintain contact there must be a centripetal acceleration. You'll also need conservation of energy.

 

[Bonulo]

Yup. I think I've got it now, I sat with a different study group after the first one went home, and reached the following:

2 b) U1 + K1 = U2 + K2, solving for v2^2 (the final speed squared). Then ma=m*v2^2/R = mg gives another expression for v2^2. This gives v1^2 = 5gR, as the square of the needed initial speed to reach the hoop top.

2 c) Same conditions - I get Theta = 19,47 degrees.