What Determines the Normal Force on a Ball in a Loop-the-Loop?

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SUMMARY

The discussion focuses on calculating the normal force acting on a 150-gram ball at the lowest point of a circular loop with a radius of 20 centimeters. The ball starts with an initial speed of 3 m/s from a height of 80 centimeters. The relevant equation used is n - mg = m(v²/r), where n is the normal force, m is mass, g is gravitational acceleration (9.8 m/s²), and r is the radius of the loop. The correct approach involves applying the conservation of energy to determine the velocity at the bottom of the loop, leading to an accurate calculation of the normal force.

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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.
PHYSICS 53 HWK 6 help Q1 figure.jpg


Homework Equations


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

Would I use v=sqrt(2gh) in this example?
 
fantisism said:
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.
 
Right.I didn't read the initial velocity part in the problem.
 
atom jana said:
The initial kinetic energy as well as the gravitational potential energy should be taken into account.
cnh1995 said:
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?
 
The expression would be 0.5*m*vf2-0.5*m*vi2=mgh.
 
cnh1995 said:
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?
 
  • #10
fantisism said:
So if I had to solve vf, it would be sqrt(2gh)+3 m/s?
No.
 
  • #11
fantisism said:
it would be sqrt(2gh)+3 m/s?
No. Rearrange the terms carefully.
 
  • #12
ehild said:
No.

cnh1995 said:
No. Rearrange the terms carefully.
vf=sqrt(2gh+(vi)2)?
 
  • #13
fantisism said:
vf=sqrt(2gh+(vi)2)?
Looks good!
 
  • #14
cnh1995 said:
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)?
 
  • #15
fantisism said:
I would subtract vi from vf
Why?
fantisism said:
n-mg=m(v2/r)?
v in this formula is nothing but your vf.
 
  • #16
cnh1995 said:
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!
 

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