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

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Homework Help Overview

The problem involves a ball sliding down a frictionless track into a circular loop, where the task is to determine the normal force acting on the ball at the lowest point of the loop. The scenario includes considerations of initial speed and height, as well as the conversion of potential energy to kinetic energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles to find the ball's speed at the bottom of the loop. There are questions about the correct use of formulas and the interpretation of initial conditions, particularly regarding kinetic and potential energy.

Discussion Status

Participants are actively engaging with the problem, offering hints and corrections regarding the energy conservation approach. There is a focus on clarifying the relationship between initial and final velocities, and how these relate to the normal force calculation. No consensus has been reached yet, but there are productive exchanges about the necessary equations.

Contextual Notes

Some participants note the importance of considering both initial kinetic energy and gravitational potential energy in the calculations. There is also a recognition of the need to carefully rearrange terms in the equations being used.

fantisism
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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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