Roller Coaster/Energy conservation

[Jordanosaur]

Hi All -

Here's the problem I have:

A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R. What minimum height h
must the block have to make a full run around the loop without falling off? The answer is to be given as a multiple of R.

I think that this problem should be easy, but I guess I'm not seeing the right angle of approach yet. I know that the minimum speed for the coaster to make a full run would be sqrt(Rg), and I think that the solution could be obtained though KE (1/2mv^2) and PE(mgh) laws. However, I can't see the right place to start on this one. If anyone can give me a pointer in the right direction, I would appreciate it. I am certainly not looking for a free solution, just a hint or two would be great.

Thanks

Jordan

 

[Dick]

You are exactly right that you want to use conservation laws. Here's one hint. The speed of the block when it reaches the top of the loop has to have a minimum value. Its radial acceleration around the loop has to equal the acceleration of gravity or it will fall off. What's a formula for the acceleration of a body in circular motion? Add that to your conservation equations.

 

[Jordanosaur]

Thanks for your response!

So the radial acceleration in circular motion is (v^2/R), and when it is set equal to the acceleration of gravity, you can conclude that Vmin = sqrt (gR).

Now, I'm thinking that we need to know the velocity at the bottom of the ramp entering the loop, and we can find the height of the ramp from that:

1/2m(sqrt(gr))^2 = mgh

= 1/2mgr = mgh

mg cancels, and we arrive at h = 1/2 R, but I know that's not right as the answer is supposed to be 2.5R.

This the first problem I have tried to use conservation laws related to circular motion, and I'm definitely missing a vital part of theory that applies to this type of problem - I know that the accel. in the loop must equal gravity, and that the height of the ramp is responsible for accruing the minimum loop-entrance velocity to allow a full revolution.

Any further suggestions?

Thanks again

 

[Dick]

It doesn't have just kinetic energy at the top of the loop. It needs extra gravitational potential energy to climb from the bottom of the loop to the top. How much? Then put it in here:

1/2m(sqrt(gr))^2+PE = mgh

 

[Jordanosaur]

Beautiful -

That's exactly what i was missing - It totally makes sense as the PE is increasing as the KE goes down toward the top of the coaster. That's what I didn't think to factor into the final equation.

I really appreciate your help, and you have saved me a couple extra hours of head scratching.

Jordan