# Velocity and acceleration in a frictionless circular loop

1. Aug 4, 2013

### KKiefhaber

This is a problem I encountered while studying/practicing for my upcoming MCAT exam:

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

A 1kg block slides down a ramp and then around a circular loop of radius 10m. Assume that all surfaces are frictionless.

1)What is the minimum height of the ramp required so that the block makes it all the way around the loop without falling? (solved)

2)How fast does the block need to be going at the bottom of the ramp so that the acceleration of the block at the top of the loop is 4g? (stuck)

2. Relevant equations

$$U=mgh$$
$$KE={1}/{2}mv^2$$
$$F_(net)=ma_y=mg+F_N$$

3. The attempt at a solution

In the initial problem, I set $$F_N=0$$ because that would be the point at which the block would begin to fall off the loop, allowing me to find the minimum height. Using $$a_c$$ for $$a_y$$ I got that $$v=\sqrt{gr}.$$
Using energy equations to set the potential energy at the top of the track equal to the (potential energy+kinetic energy) found at the top of the loop, I found that the minimum height of the ramp would be 25m.

However, for problem 2, the solutions I have tried have been incorrect.
Given that the desired acceleration of the block is $$4g$$ I first set $$4g=\frac{v^2}{r}$$
This gives v=19.95m/s which is incorrect according to the answer key provided (v=28m/s).

Next I attempted setting $$ma_c=mg+F_N$$ or $$\frac{mv^2}{r} = mg+F_N$$
but couldn't get much further.

A third attempt at using energy relationships led me to
$$\frac{1}{2}mv_1^2=mg2r+\frac{1}{2}mv_2^2$$ where the left side is the kinetic energy at the bottom of the ramp just before the loop, and the right side is the total energy (kinetic and potential) at the top of the loop. After cancelling out $$m$$ I was again unable to proceed.

Thank you very much for taking the time to go through all of this, I look forward to any help or pointing in the right direction anyone can give me.

2. Aug 4, 2013

### voko

In the first attempt, you found the velocity at the top of the loop. But you need to find the velocity at the bottom.

In your third attempt, you found the correct equation relating these two velocities.

You are almost there.

3. Aug 4, 2013

### KKiefhaber

Thank you! For some reason, spelling it out that way made it click!

I didn't realize that the initial velocity I'd solved for was actually the velocity at the top of the loop, not just an incorrect answer. I used it for my v sub 2 which allowed me to solve my energy equation for v sub 1 giving me 28.1 as the answer.

I was so frustrated, I didn't realize I'd actually solved for part of the problem :)

Thanks again!