# Energy conservation problem

1. Apr 15, 2012

### Jalo

1. The problem statement, all variables and given/known data

A particle of mass m starts his motion from rest in the top of a inclined plane. It slides down the plane and enters a circular rail of radius R.
(a) What's the minimum value of b for the particle not to
Uma partícula de massa m parte do repouso no topo de um plano inclinado,
desliza e entra numa calha circular de raio R como mostra a figura.
(a) What is the minimum value of b so that the particle does not leave the rail in
her movement?
(b)Assume that b = R. Determine the force exerted on the particle by the rail
when it reaches the height R.

2. Relevant equations

Em=K+U

Em is mechanic energy, K is kinectic energy and U is gravitic potential energy.

3. The attempt at a solution

(a)When the particle starts his motion her mechanic energy is mg(2R+b). When it reaches the top of the circular rail her mechanic energy will be mg(2R)+.5mv2.
Since the mechanic energy is a constant:

mg(2R+b)=mg(2R)+.5*mv2
Solving the equation I got b=-v2^2/(2*g)

I'm not sure if I did it the right way tho.

(b) I have no idea how to solve it. I tried working the function I got in (a) and derivating to obtain the accelaration, without success, since the accelaration would be 0. I need some hint to solve it...

Thanks.

2. Apr 15, 2012

### billVancouver

Hi Jalo,

First let's double-check part (a). You start off really well, by noticing that the gravitational potential energy of the particle initially is PE = mg*(2R+b). Since the particle starts at rest, this is its total energy for the whole problem, due to conservation of energy and since there doesn't appear to be any friction or drag considered in the problem. Therefore, what is the particle's kinetic energy at the top of the circular rail? It should be the total energy you wrote down, minus the potential energy at a height of 2R (the height that the particle is at at the top of the rail). Now that you know the particle's kinetic energy, you can find its velocity, and with the velocity you can determine the centripetal acceleration needed to keep it moving in a circle of radius R. Equate that force to the force of gravity, and you should get the correct answer for (a).

For (b), you can do pretty much the same algebra - you can find the particle's kinetic energy when it gets to a height R via conservation of energy just like above, which will tell you a velocity and therefore the necessary centripetal acceleration needed to keep the particle traveling in a circle of radius R. Finally, this acceleration can be translated into the necessary force via one of Newton's laws.

Hope this helps,
Bill Mills

Last edited by a moderator: Apr 16, 2012
3. Apr 16, 2012

### MrWarlock616

The problem is similar to one I had posted earlier. (See here). The (b) part is very simple. The centripetal force is what they want. It is given by $F_c = \frac{m v^2}{r}$
But at the top of a circle, $v^2 = r g$

$\therefore F_c = \frac{m r g}{r}=m g$

The force acting at height R is thus $mg$, its weight.