Conservation of Energy -- Object around a loop

[PhysicsCollegeGirl]

Hello!
I am having problems with this exercise if someone can please help me.

Homework Statement

In order to go all the way around a frictionless circular loop of R=2.5 m , how fast must a mass be moving as it exits at the bottom of the loop.

2. The attempt at a solution
I am doing this problem in two parts. First I find the velocity at the top, and then I look for the velocity at the bottom. So,

part 1. V at top
W + PEi + KEi = KEf + PEf + Heat loss
0 + 0 + (1/2)(m)(v^2) = 0 + mgh + 0
v= sqrt [(gh)(2)] = sqrt [(9.8*4*R)] = 9.89

part 2. V at bottom
W + PEi + KEi = KEf + PEf + Heat loss
0 + mgh + (1/2)(m)(v_top^2) = (1/2)(m)(v_bot^2) + 0 + 0
sqrt[((gh)+(1/2)(v_top^2))*2] = v_bot
sqrt[((g2*R)+(1/2)(v_top^2))*2] = v_bot
v_bot = 11.06
However, the homework says I should be getting 12 m/s as a response.

 

[TomHart]

sqrt[((g2*R)+(1/2)(vtop^2))*2] = vbot

You may want to check your math. It looks like 11.06 is wrong using your equation - unless I did the math wrong, which I have been known to do on (Edit: many) occasions.

 

[TomHart]

Wait a minute. I went back and looked at your "velocity at the top" calculation. I think what the problem is asking is this: What is the minimum exit speed for the mass to maintain contact with the surface for the entire loop. So what that means is that the mass is just barely making contact with the surface at the top. In other words, the normal force from the track on the mass is 0 at the top of the loop. So how do you calculate the speed of the mass at the top under that condition?

 

[haruspex]

a frictionless circular loop

This is unclear. Is it inside the loop, with nothing to prevent it falling off in the top half, or is it like a bead on a wire?

exits at the bottom of the loop.

Enters?

0 + 0 + (1/2)(m)(v^2) = 0 + mgh + 0

What is the basis of this equation? You seem to be finding the KE after descending h from somewhere above the top of the loop.

 

[CWatters]

I also got 11.07m/s.

I assumed it was a loop that you could fall off. I used a Free Body Diagram to analyse the forces and calculate the minimum velocity at the top needed to stick to the inside of a loop. That worked out at 4.95m/s. Then applied conservation of energy to get the velocity at the bottom = 11.07m/s.

You get the same value if you assume the problem was asking for the entry or exit velocity at the bottom because it's frictionless.

If you assume it's a bead on a wire (rather than a track you can fall off) you get a lower value than 11.07m/s because the mass "only just" has to reach the top.

I think the book answer is wrong.

 

[TomHart]

I also got 11.07m/s.

But @CWatters, she got a different minimum velocity (9.89 m/s) at the top than you did so she should not have gotten the same final answer. Please take a look at her method of calculating the velocity at the top. Thank you.

 

[TomHart]

You seem to be finding the KE after descending h from somewhere above the top of the loop.

@haruspex, I just now saw this in your post. I like how you worded that.

 

[CWatters]

part 1. V at top
W + PEi + KEi = KEf + PEf + Heat loss
0 + 0 + (1/2)(m)(v^2) = 0 + mgh + 0
v= sqrt [(gh)(2)] = sqrt [(9.8*4*R)] = 9.89

The above calculation equates the initial Kinetic Energy (KEi) with the final Potential Energy PEf. So the velocity you are calculating is not the velocity at the top. You have calculated the velocity the mass would need at the bottom in order to coast up to the top and arrive there with zero velocity (KEf=0). That's not sufficient velocity to avoid the mass falling off the track (unless it's a bead on a wire type).

 

[PhysicsCollegeGirl]

Wait a minute. I went back and looked at your "velocity at the top" calculation. I think what the problem is asking is this: What is the minimum exit speed for the mass to maintain contact with the surface for the entire loop. So what that means is that the mass is just barely making contact with the surface at the top. In other words, the normal force from the track on the mass is 0 at the top of the loop. So how do you calculate the speed of the mass at the top under that condition?

This is unclear. Is it inside the loop, with nothing to prevent it falling off in the top half, or is it like a bead on a wire?

Enters?

What is the basis of this equation? You seem to be finding the KE after descending h from somewhere above the top of the loop.

I also got 11.07m/s.

I assumed it was a loop that you could fall off. I used a Free Body Diagram to analyse the forces and calculate the minimum velocity at the top needed to stick to the inside of a loop. That worked out at 4.95m/s. Then applied conservation of energy to get the velocity at the bottom = 11.07m/s.

You get the same value if you assume the problem was asking for the entry or exit velocity at the bottom because it's frictionless.

If you assume it's a bead on a wire (rather than a track you can fall off) you get a lower value than 11.07m/s because the mass "only just" has to reach the top.

I think the book answer is wrong.

The above calculation equates the initial Kinetic Energy (KEi) with the final Potential Energy PEf. So the velocity you are calculating is not the velocity at the top. You have calculated the velocity the mass would need at the bottom in order to coast up to the top and arrive there with zero velocity (KEf=0). That's not sufficient velocity to avoid the mass falling off the track (unless it's a bead on a wire type).

Ok, I understand now what you are saying. It is indeed a problem with a mass that can fall off, like a roller coaster, I didn't put the picture but it is like that. Using net force = mass * centripetal acceleration I got 4.95 m/s, just like @CWatters.
Thank you for making it more clear.
With that info, I looked for the V_bot, but I still got 11.06 m/s. But I am pretty sure that is the right answer anyway.
Sadly, this was a question on a multiple choice test and the options they gave you were: a. 8 m/s b. 10 m/s and c. 12 m/s
But I understand it much better now, so thank you all!

 

[haruspex]

You get the same value if you assume the problem was asking for the entry or exit velocity

Sure, but it is odd to use "must" in that case. It defies everyday ways of expressing causality, which would ask what speed it must have on entry, or what speed it will have on exit.

Given the answer 12m/s, the radius should have been 3m, or nearly.

 

[TomHart]

Sure, but it is odd to use "must" in that case. It defies everyday ways of expressing causality, which would ask what speed it must have on entry, or what speed it will have on exit.

I like how you worded that too. The wording of the problem caused confusion.