Homework Help: Object sliding through a loop

1. Mar 13, 2015

Ishaan S

1. The problem statement, all variables and given/known data
A small mass m slides without friction along the looped apparatus shown in the attached file. (a) If the object is to remain on the track, even at the top of the circle(whose radius is r), from what minimum height hm must it be released(from rest)? (b) If it is released at a height h = 0.80hm, at what height will it lose contact with the track?

2. Relevant equations
ac = v2/r

KE = .5mv2

PEgrav = mgh

3. The attempt at a solution
For a:

I knew that for the object to complete the loop, it must be able to complete the extreme case: the top. I knew that the centripetal acceleration must be g. Therefore, from the centripetal acceleration formula, the speed had to be at least v = √(rg).

From the law of the conservation of energy, I equated the initial and final(at the top of the loop) energies:

mghm = 2rmg + .5mv2

I cancelled out the m and substituted v from the centripetal acceleration equation and got

ghm = 2rg + .5rg = 2.5rg

Cancel out g

hm = 2.5r

For b:

The initial height .80hm = 2r.

Using the law of the conservation of energy

2rmg = mgh + .5m(0)2.

I substituted 0 for v in the equation above because I thought that the velocity had to be 0 when the object left the track.

Solving for h, I got h = 2r.

I have no idea if this is right. Could someone please tell me if this is right, and if not, give me a nudge in the right direction.

Thanks.

Attached Files:

• 20150313_163507 (1).jpg
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Last edited: Mar 13, 2015
2. Mar 13, 2015

TSny

Hello, Ishaan.

Your work for part (a) looks very good!

For part (b) your assumption that v = 0 when the mass leaves the track is not correct. Hint: What force on the mass must go to zero at the instant the mass leaves the track?

3. Mar 13, 2015

Ishaan S

Ok.

So I set up an equation.

Fnormal + mgcosθ = mv2/r

where θ is the angle between the normal force and the gravitational force.

I set Fnormal equal to zero because when the object leaves the track, the normal force would be 0.

mgcosθ = mv2/r

I solved for v2.

v2 = rgcosθ

I then set up the conservation of energy equation.

2rmg = mgh + mv2/2.

I substituted rgcosθ for v2 due to my previous derivation. I then cancelled mg from both sides.

2r = h + r(cosθ)/2,

Then I set up a diagram that I have attached.

I solved for cosθ using the diagram and substituted it into the above equation.

2r = h + (h-r)/2.

I solved for h and got

h = (5/3)r.

Is this correct?

Thanks.

Attached Files:

• 20150313_175914.jpg
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Views:
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4. Mar 13, 2015

TSny

That all looks correct. Very nice work!