# Loop-the-loop ball Physics question

## Homework Statement

A 150 gram ball slides down a smooth track with no friction, which has a circular loop with radius 20 centimeters. The ball has an initial speed of 3 m/s. Its initial position is 80 centimeters above the lowest point of the circular loop. Calculate the normal force of the track on the ball when it is at the lowest point of the circular loop. n-mg=m(v2/r)

## The Attempt at a Solution

I have drawn a force diagram for when the ball is at the lowest point on the circular loop with n (normal force) pointing upwards and mg downwards. I THINK the only formula I need for this problem is what I already mentioned. I'm sure I substituted in most of the values right except for the v. I made v equal to 3 m/s but I'm sure it'll be faster when it's at the bottom of the circular loop. With v=3 m/s, m=0.15 kg, r=0.2 m, and g=9.8 m/s2, I got n to be equal to 8.22 N.

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cnh1995
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Gold Member
I'm sure it'll be faster when it's at the bottom of the circular loop
Yes.
Hint: Conservetion of energy...

Yes.
Hint: Conservetion of energy...
Would I use v=sqrt(2gh) in this example?

cnh1995
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Gold Member
Would I use v=sqrt(2gh) in this example?
Right. Lost potential energy is converted into kinetic energy.

The initial kinetic energy as well as the gravitational potential energy should be taken into account.

cnh1995
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Right.I didn't read the initial velocity part in the problem.

The initial kinetic energy as well as the gravitational potential energy should be taken into account.
Right.I didn't read the initial velocity part in the problem.
So in order to find the velocity when the ball reaches the bottom of the loop, I would use 0.5*m*(vi)2+0.5*m*(vf)2=mgh and solve for vf?

cnh1995
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The expression would be 0.5*m*vf2-0.5*m*vi2=mgh.

The expression would be 0.5*m*vf2-0.5*m*vi2=mgh.
So if I had to solve vf, it would be sqrt(2gh)+3 m/s?

ehild
Homework Helper
So if I had to solve vf, it would be sqrt(2gh)+3 m/s?
No.

cnh1995
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Gold Member
it would be sqrt(2gh)+3 m/s?
No. Rearrange the terms carefully.

No.
No. Rearrange the terms carefully.
vf=sqrt(2gh+(vi)2)?

cnh1995
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Gold Member
vf=sqrt(2gh+(vi)2)?
Looks good!

Looks good!
And then once I find that out, I would subtract vi from vf and use that value for v in n-mg=m(v2/r)?

cnh1995
Homework Helper
Gold Member
I would subtract vi from vf
Why?
n-mg=m(v2/r)?
v in this formula is nothing but your vf.

Why?

v in this formula is nothing but your vf.
Oh alright then. I thought v in the n-mg=m(v2/r) was the change in velocity, although now that I think about it, it's unnecessary. Thank you so much!