Potential Energy and Loop-de-Loops

[brinstar]

Homework Statement

A bead slides without friction around a loop–the–loop (see figure below). The bead is released from rest at a heighth = 3.60R.

(a) What is its speed at point

? (Use the following as necessary: the acceleration due to gravity g, and R.)
v = _________________


(b) How large is the normal force on the bead at point

if its mass is 4.70 grams?
Find the magnitude and direction.

Homework Equations

u = mgh
k = .5mv^2
ui + ki = uf + kf
g = 9.8 m/s^2
h = 3.60R

The Attempt at a Solution

For a, I don't know how much farther I can go. the farthest I've gotten is this:

where h = 3.60R

ui + ki = uf + kf
mgh + 0 = .5mv^2 + mgh
gh = .5v^2 + gh

If I subtract gh from the right though, I'll get 0... which is obviously not the answer here. So I need major help there.

As for b, I know the direction is downward. It's upside down so if I flip the entire diagram, I see that normal force is really pointing down.

As for the first part, I just converted grams to kg, then multiplied by 9.8 m/s^2. However, the website I'm entering the answer into keeps saying I'm wrong when I type in when I input my answer 0.04606 N, so I'm not sure what's up there... I also tried 0.0461 N, and still no dice...

Thanks for any help!

 

[haruspex]

ui + ki = uf + kf
mgh + 0 = .5mv^2 + mgh

What does the variable uf represent? What is the 'final' state in the calculation you are trying to do? (I.e., where is the bead at the end?)

 

[brinstar]

What does the variable uf represent? What is the 'final' state in the calculation you are trying to do? (I.e., where is the bead at the end?)

uf is final potential energy, which I'm counting as the top of the loop at point A. So, although I don't know if the set up is correct, I'm making the final stop to be at the loop-de-loop point A.

 

[haruspex]

uf is final potential energy, which I'm counting as the top of the loop at point A. So, although I don't know if the set up is correct, I'm making the final stop to be at the loop-de-loop point A.

Ok, so what should uf equal? Why have you substituted mgh?

 

[brinstar]

Ok, so what should uf equal? Why have you substituted mgh?

where (initial height) = 3.60R

and I fixed the initial h and h at a, as I just realized how bad of a mistake that was

ui + ki = kf + uf
mg(initial height) + 0 = .5mv^2 + mg(height at a)
3.60Rg = .5v^2 + g(height at a)

I don't know how to get height at A based on the 3.60R stuff so I am totally lost there. My book keeps saying that height at a is 2R, but I don't understand that conceptually and I don't just want to put down something to answer the question without fully understanding it :(

 

[haruspex]

where (initial height) = 3.60R

and I fixed the initial h and h at a, as I just realized how bad of a mistake that was

ui + ki = kf + uf
mg(initial height) + 0 = .5mv^2 + mg(height at a)
3.60Rg = .5v^2 + g(height at a)

I don't know how to get height at A based on the 3.60R stuff so I am totally lost there. My book keeps saying that height at a is 2R, but I don't understand that conceptually and I don't just want to put down something to answer the question without fully understanding it :(

For the purposes of gravitational PE, you have to choose a common baseline. The initial height h is measured from the bottom of the loop,so the bottom of the loop is the baseline. The loop has radius R, so what is the height of the top of the loop?

 

[brinstar]

For the purposes of gravitational PE, you have to choose a common baseline. The initial height h is measured from the bottom of the loop,so the bottom of the loop is the baseline. The loop has radius R, so what is the height of the top of the loop?

OH MY G THAT MAKES SO MUCH MORE SENSE!

Thank you!

The height is 2R, as the diameter is the height of the loop.

3.60Rg = .5v^2 + g2R
3.60Rg - 2Rg = .5v^2
2Rg(3.60-2) = v^2
v = sqrt(Rg(3.20))

So then, proceeding on

Fn + Fg = sum of forces = ma

So would ma be centripetal acceleration?

 

[haruspex]

So would ma be centripetal acceleration?

ma would be the centripetal force, a the centripetal acceleration, which I assume is what you meant.

 

[brinstar]

ma would be the centripetal force, a the centripetal acceleration, which I assume is what you meant.

Yes, thank you, that's what i meant

Fn + Fg = sum of forces = ma = mv^2/R

Fn + Fg = m (3.2gR)/R
Fn + Fg = 3.2gm
Fn + mg = 3.2mg
Fn = 3.2mg - 1mg
Fn = mg(2.2)
Fn = (4.70 x 10^-3)(9.8)(2.2)
Fn = 0.101332 N

Okay, I hope I got it now!

 

[haruspex]

Yes, thank you, that's what i meant

Fn + Fg = sum of forces = ma = mv^2/R

Fn + Fg = m (3.2gR)/R
Fn + Fg = 3.2gm
Fn + mg = 3.2mg
Fn = 3.2mg - 1mg
Fn = mg(2.2)
Fn = (4.70 x 10^-3)(9.8)(2.2)
Fn = 0.101332 N

Okay, I hope I got it now!

That looks right.

 

[brinstar]

That looks right.

Awesome, thanks so much!