Mass slides on loop-the-loop track, find original height.

[oddjobmj]

Homework Statement

http://imgur.com/b7y3Khb

A small cart slides without friction on the loop-the-loop track, as shown, starting from rest at initial height h. The weight of the cart is W. When the cart is at the point A --- the top of the loop --- the force on the track is 0.6 W. Then what was the initial height h?
Data: W = 111.0 N; R = 16.2 m.

Homework Equations

E_k=$\frac{1}{2}$mv²
E_u=mgh
Centripetal force = $\frac{mv^2}{r}$

The Attempt at a Solution

If I can find the kinetic energy at the given point I can then solve for v. With v, which contains h, I can use the equation for centripetal force to solve for h.

m=w
I know the original potential energy = wgh

The potential at the point of interest is = 2Rwg

Thus the kinetic energy = (original)-(current) = wgh-2Rwg = mg(h-2R) = $\frac{1}{2}$wv²

Thus v=$\sqrt{2g(h-2R)}$

Plugging v into the equation for centripetal force I get

Force Centripetal = $\frac{2gw(h-2R)}{r}$

I think this is where I am messing up:

The force on the object is equal to 0.6w and the only two forces acting on the object are the loop and gravity.

0.6w= [Force Centripetal]-[Force Gravity] ?

F_c=0.6w+F_g => $\frac{2gw(h-2R)}{r}$=.6w+wg

Solving that for h:

h=$\frac{3(5gR+R)}{10g}$

Plugging in the known values I get h=24.8m which is wrong. Any suggestions?

Thank you!

 

[TSny]

m=w

Mass and weight are different concepts.

Note how in your final answer: h=$\frac{3(5gR+R)}{10g}$ you have 5gR added to R. But these quantities have different dimensions (or units). This shows something is wrong and it's due to taking m = w.

Otherwise, your work looks good.

 

[collinsmark]

Homework Statement

http://imgur.com/b7y3Khb

A small cart slides without friction on the loop-the-loop track, as shown, starting from rest at initial height h. The weight of the cart is W. When the cart is at the point A --- the top of the loop --- the force on the track is 0.6 W. Then what was the initial height h?
Data: W = 111.0 N; R = 16.2 m.

Homework Equations

E_k=$\frac{1}{2}$mv²
E_u=mgh
Centripetal force = $\frac{mv^2}{r}$

The Attempt at a Solution

If I can find the kinetic energy at the given point I can then solve for v. With v, which contains h, I can use the equation for centripetal force to solve for h.

m=w

I'm not following you there. If w is the weight, then w = mg.

I know the original potential energy = wgh

You threw an extra g in there, but g is included in the weight. It's not necessary to put it in there again.

original potential energy = wh

The potential at the point of interest is = 2Rwg

2Rw

Thus the kinetic energy = (original)-(current) = wgh-2Rwg = mg(h-2R) = $\frac{1}{2}$wv²

I would say,

w(h - 2R) = ½(w/g)v².

Thus v=$\sqrt{2g(h-2R)}$

Your expression for v is correct.

Plugging v into the equation for centripetal force I get

Force Centripetal = $\frac{2gw(h-2R)}{r}$

You have an extra g in there.

I think this is where I am messing up:

The force on the object is equal to 0.6w and the only two forces acting on the object are the loop and gravity.

0.6w= [Force Centripetal]-[Force Gravity] ?

F_c=0.6w+F_g => $\frac{2gw(h-2R)}{r}$=.6w+wg

And that's where things make a difference, when treating weight as mass.

So far you've been equating mass and weight, and it comes to bite back here. If you are treating w as mass, then 0.6w is also a mass, and wg is a force. You're mixing mass and force.

Instead, treat w, the weight, as a force; don't treat it as a mass.

That gives you, on the right hand side of the equation, 0.6w + w

I'll let you take it from there.

[Edit: TSny beat me to the response.]