Dynamics - particle moves on smooth inside surface of hemisphere?

  • Thread starter Kate2010
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  • #1
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Homework Statement


I'm currently working on this question:

A particle moves on the smooth inside surface of the hemisphere
z = -(a^2 - r^2)^(1/2), r <= a,
where (r, theta, z) denote cylindrical polar coordinates, with the z-axis vertically upward.

Initially the particle is at z = 0, and it is projected with speed V in the theta-direction.

1. Show that the particle moves between two heights in the subsequent motion, and find them.

2. Show, too, that if the parameter b = (V^2)/4ga is very large then the difference between the two heights is approximately a/(2b).

Homework Equations



Conservation of energy 1/2 m(r'^2 + r^2 theta'^2 + z'^2) + mgz = constant
rtheta' = V
r^2 theta = aV
theta ' = aV/r^2
( ' = dot, differentiation)


The Attempt at a Solution



I'm really stuck on question 1, I got to an answer using conservation of energy that ended with solving a quadratic, this is the particle moves between z=0 and z= (V^2 + sqrt(V^4 + 16g^2a^2))/4g but when I sub in the parameter in (2) I get to the wrong answer, ie 2ab. So I guess I've done 1 wrong. Any ideas guys?
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi Kate2010! Welcome to PF! :smile:

(have a theta: θ and a square-root: √ and try using the X2 tag just above the Reply box :wink:)
2. Show, too, that if the parameter b = (V^2)/4ga is very large then the difference between the two heights is approximately a/(2b).

the particle moves between z=0 and z= (V^2 + sqrt(V^4 + 16g^2a^2))/4g but when I sub in the parameter in (2) I get to the wrong answer, ie 2ab. So I guess I've done 1 wrong. Any ideas guys?
How did you get 2ab from that equation (I get a/2b)?

(Did you use √(1 + X) ~ 1 + X/2 ?)
 
  • #3
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I can get to a(b + √(b² +1))

Then I said √(b² +1) is roughly equal to b, I think this is where I'm going wrong?
 
  • #4
tiny-tim
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oops! :redface: i've just noticed that when i got a/2b, i misread your equation as z= (V^2 - sqrt(V^4 + 16g^2a^2))/4g (with a minus instead of a plus) …

are you sure it isn't that (because the z should be negative, shouldn't it)?
 
  • #5
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Yes you're right, z should be negative.

So I get to a(b - √(b² +1))

I still don't understand how this is approximately equal to a/2b.
 
  • #6
tiny-tim
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Yes you're right, z should be negative.
he he :biggrin: … i'm accidentally right again! :redface:
So I get to a(b - √(b² +1))

I still don't understand how this is approximately equal to a/2b.
b is large, so you need to divide by b to get (1 + something-small) inside the √ …

a(b - √(b² +1)) = ab(1 - √(1 +1/b²)).

Now just use √(1 + X) ~ 1 + X/2. :wink:
 
  • #7
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Got it! Thanks a lot :biggrin:
 

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