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Marble around the loop-the-loop

  • Thread starter Lydia22
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  • #1
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Homework Statement




A marble rolls down a track and around a loop-the-loop of radius R. The marble has mass m and radius r. What minimum height h must the track have for the marble to make it around the loop-the-loop without falling off?


Homework Equations


mgh = 1/2mv^2 + 1/2Iw^2


The Attempt at a Solution



mgh= 1/2mv^2 + 1/2mr^2*v^2/r^2

gh=1/2v^2 +v^2
h=3/*2(v^2)/g

At the top of the loop
n+mg=mv^2 /r
it just need the minimum velocity therefore the normal will be zero
v=√gr
h=3/2*r which is incorrect please help
 

Answers and Replies

  • #2
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mgh= 1/2mv^2 + 1/2mr^2*v^2/r^2
Is a marble rolling through the track or is a hoop rolling through?

What is the moment of inertia of a marble? :wink:
 
  • #3
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mgh= 1/2mv^2 + 2/5mr^2*v^2/r^2

gh=1/2v^2 +1/5v^2
h=1/10(v^2)/g

At the top of the loop
n+mg=mv^2 /r
it just need the minimum velocity therefore the normal will be zero
v=√gr
h=1/10g
The marble is rolling in a hoop
 
  • #4
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mgh= 1/2mv^2 + 2/5mr^2*v^2/r^2

gh=1/2v^2 +1/5v^2
h=1/10(v^2)/g
You'll have to double check your math. :smile: 1/2 + 1/5 isn't 1/10
At the top of the loop
n+mg=mv^2 /r
it just need the minimum velocity therefore the normal will be zero
v=√gr
h=1/10g
The marble is rolling in a hoop
The way you are approaching the problem is a good approximation if R >> r.

I'm not sure how precise you are supposed to get for this problem. But if you want to get technically exact, the radius of the loop-the-loop that the ball moves through not quite R. The marble has its own radius, meaning that the radius traversed by the ball's center of mass through the loop-the-loop is only R - r.

If r << R, this differences is negligible. But if you have a big marble and a small loop-the-loop it can be significant.
 
  • #5
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And there's something else that's very important too. The problem statement says,
What minimum height h must the track have
Is h relative to the ground or to the top of the loop (i.e. 2R)? This makes a big difference.
 
  • #6
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gh = 7/10 [v]^2
[v]^2=10/7 gh

v=√gR-r
10/7gh =g(R-r)

h=10/7(R-r)
 
  • #7
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the height "h" is relative to the ground
 
  • #8
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gh = 7/10 [v]^2
[v]^2=10/7 gh

v=√gR-r
10/7gh =g(R-r)

h=10/7(R-r)
By the way, I'm assuming that your above equation is [itex] h = \frac{7}{10}(R-r) [/itex].

Okay, that's the right idea, but it's making an assumption about h.

Your original formula, [itex] mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 [/itex], assumes that all of the gravitational potential energy mgh gets completely converted into kinetic energy.

The assumption your method makes is that h is relative to the center of the ball when it is at the top of the loop-the-loop.
the height "h" is relative to the ground
In that case, you're not quite finished yet.

When the ball is at the top of the loop-the-loop (at a height of 2(R - r)) does it have any gravitational potential energy?
 
  • #9
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yes it will have gravitational potential energy ,so do i add the radius of the marble to the height?
 
  • #10
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yes it will have gravitational potential energy ,so do i add the radius of the marble to the height?
You've done most of the work already. :approve: So you're almost there.

The way you've treated h so far, h is the height above the ball when the ball is at the top of the loop-the-loop (which is 2(R - r) above the ground). That's because in your original equation, you did not leave a gravitational potential energy term on the right hand side of the equation.

So if you add those two heights together, you'll have the minimum starting height of the ball relative to the ground. :smile:

[i.e (height from ground to ball at top of loo-the-loop)+(height from ball at top of loop-the-loop to starting height above that)]
 

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