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

In summary, when a 150 gram ball slides down a frictionless track with a circular loop of radius 20 centimeters and an initial speed of 3 m/s, the normal force of the track on the ball at the lowest point of the loop can be calculated using the equation n-mg=m(v2/r). The initial kinetic energy and gravitational potential energy should be taken into account, and the velocity at the bottom of the loop can be found using the conservation of energy equation 0.5*m*vf2-0.5*m*vi2=mgh. Once the final velocity is determined, it can be substituted into the original equation to solve for the normal force.
  • #1
fantisism
16
0

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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  • #2
fantisism said:
I'm sure it'll be faster when it's at the bottom of the circular loop
Yes.
Hint: Conservetion of energy...
 
  • #3
cnh1995 said:
Yes.
Hint: Conservetion of energy...

Would I use v=sqrt(2gh) in this example?
 
  • #4
fantisism said:
Would I use v=sqrt(2gh) in this example?
Right. Lost potential energy is converted into kinetic energy.
 
  • #5
The initial kinetic energy as well as the gravitational potential energy should be taken into account.
 
  • #6
Right.I didn't read the initial velocity part in the problem.
 
  • #7
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?
 
  • #8
The expression would be 0.5*m*vf2-0.5*m*vi2=mgh.
 
  • #9
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!
 

1. What is a loop-the-loop ball?

A loop-the-loop ball is a type of toy or experiment that involves a small ball rolling around a track with a circular loop. The ball is able to complete a full loop without falling out of the track due to the forces acting on it.

2. How does a loop-the-loop ball work?

A loop-the-loop ball works by utilizing the principles of centripetal force and gravity. The ball's inertia, or resistance to change in motion, allows it to stay on the track while the centripetal force from the track keeps it moving in a circular path. Gravity also plays a role by pulling the ball towards the bottom of the track and helping it maintain its speed.

3. What factors affect the success of a loop-the-loop ball?

The success of a loop-the-loop ball can be affected by various factors such as the size of the loop, the speed and direction of the ball, and the shape and material of the track. Other factors such as air resistance and friction can also play a role in the ball's ability to complete the loop without falling out of the track.

4. How can you calculate the speed needed for a loop-the-loop ball?

The speed needed for a loop-the-loop ball can be calculated using the formula v = √(Rg), where v is the speed, R is the radius of the loop, and g is the acceleration due to gravity. This formula takes into account the centripetal force needed to keep the ball in the loop.

5. What are some real-life applications of loop-the-loop ball physics?

Loop-the-loop ball physics can be applied in various fields such as roller coaster design, amusement park rides, and even sports like skateboarding and snowboarding. It can also be used in engineering and physics experiments to demonstrate concepts such as centripetal force and gravity.

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