Calculating Horizontal Force on Ball in Loop-the-Loop Track

In summary, the ball has some radius r and some mass m. At the bottom of the ramp there is a loop. The loop has radius R. If the ball is 1/4 of the way completed the loop, what is the horizontal component of the net force on the ball. And it is also given that the ball starts up the ramp at height 6R.
  • #1
Little Dump
19
0
Ok there's this ball rolling down a ramp. The ball has some radius r and some mass m. At the bottom of the ramp there is a loop. The loop has radius R.

The question is, if the ball is 1/4 of the way completed the loop, what is the horizontal component of the net force on the ball. And it is also given that the ball starts up the ramp at height 6R.

Ok so I've been messing around, and what getting me is that when the ball starts the go up in the loop, gravity will do some work. The problem is, I do not know how to calculate the work done by gravity.

Perhaps I am doing the question wrong because I am analysing the rotation of the ball in the loop, not the actual rotation of the ball itself.

Basically what I am doing is using energy and here's my equations

mg(6R) + Wgrav = 1/2mv^2 + 1/2I(omega)^2 + mgR
and
v = R(omega)

so we make the substitution and solve for omega

then we go back and solve for v

the we do

a = v^2/R

and finally

F=ma

but I may be doing this completely wrong so any help will be great. Here is the original question with diagram:




A small solid marble of mass 12 g and radius 3.8 cm will roll without slipping along the loop-the-loop track shown in Fig. 12-33 if it is released from rest somewhere on the straight section of track. (a) From what initial height h above the bottom of the track must the marble be released if it is to be on the verge of leaving the track at the top of the loop? (The radius of the loop-the-loop is 4.0 m; Note that it is much greater than the radius of the marble.) (b) If the marble is released from height 6R above the bottom of the track, what is the horizontal component of the force acting on it at point Q?


The answer with these number is 0.840N.
 

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  • #2
I see terms in both sides for gravitational potential energy... that is the work done by gravity!

More precisely, the work done by gravity is the difference in gravitational potential energy, so you've already accounted for the work done by gravity because you've included the GPE terms... you don't need to add an additional work due to gravity term.
 
  • #3
Alright but I still don't get the right answer.

Should i maybe analyse the ball instead of the loop?


Im really lost
 
  • #4
Aha, here's your problem:

mg(6R) + Wgrav = 1/2mv^2 + 1/2I(omega)^2 + mgR
and
v = R(omega)

Those are different omegas! In the first equation, omega is the angular velocity with which the ball is rotating, and in the second equation, omega is the angular velocity with which the ball progresses around the center of the loop!


I think what you want here is:

v = r ω

Where this ω is the same as the one in your first equation and r is the radius of the ball. (I'm using the fact that the radius of the loop is much larger than the radius of the ball, thus allowing me to assume the loop is approximately flat)
 
Last edited:
  • #5
Still no good

more lost then ever

:\

Perhaps someone else has a suggestion?
 

What is the formula for calculating the horizontal force on a ball in a loop-the-loop track?

The formula for calculating the horizontal force on a ball in a loop-the-loop track is F = mv²/r, where F is the force, m is the mass of the ball, v is the velocity of the ball, and r is the radius of the loop.

How does the mass of the ball affect the horizontal force?

The mass of the ball directly affects the horizontal force. The greater the mass of the ball, the greater the force needed to keep it moving in the loop-the-loop track. This is because the force is directly proportional to the mass in the formula F = mv²/r.

Can the horizontal force be greater than the vertical force in a loop-the-loop track?

Yes, it is possible for the horizontal force to be greater than the vertical force in a loop-the-loop track. This can occur if the velocity of the ball is high enough or if the radius of the loop is small enough. However, it is important to note that in order for the ball to successfully complete the loop-the-loop, the vertical force must still be greater than or equal to the force of gravity.

How does the velocity of the ball affect the horizontal force in a loop-the-loop track?

The velocity of the ball is directly proportional to the horizontal force in a loop-the-loop track. This means that the faster the ball is moving, the greater the horizontal force needed to keep it in the loop. Additionally, a higher velocity can also result in a greater horizontal force than vertical force, which can cause the ball to fall out of the loop.

How does the radius of the loop affect the horizontal force?

The radius of the loop is inversely proportional to the horizontal force in a loop-the-loop track. This means that the larger the radius, the smaller the horizontal force needed to keep the ball in the loop. Conversely, a smaller radius will require a greater horizontal force to keep the ball in the loop.

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