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USAPhO 2008 F=ma exam #18, (gravitational potential)

  1. Jan 25, 2014 #1
    http://www.aapt.org/Programs/contest...08_fnet_ma.pdf [Broken]
    1. The problem statement, all variables and given/known data

    A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ring
    at a distance much greater than R and allowed to fall towards the ring under the influence of the
    ring’s gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same
    (linear) mass density but radius 2R, and the experiment is repeated. What is the new maximum
    speed of the particle?

    2. Relevant equations

    U = -Gm1m2 / r2

    basic kinematics

    ρ= m/V = m / (2∏rA) (A = cross sectional area, although it is negligible I believe b/c they give you linear mass density)

    mring = (2∏r)ρA

    acceleration = Fg / mparticle = (Gmring)/d2 = (G(2∏r)ρA) / d2

    3. The attempt at a solution

    Here's how I solved this problem TWO DIFFERENT WAYS (once with kinematics, once with conservation of energy) and got (√2)v both times. The answer is 2v.

    1) kinematics

    asecond instance / afirst instance = a2 / a1 = [Gm2 / x2 ] / [Gm1 / x2 ] = m2 / m1 = 4∏rAρ / 2∏rAρ = 2.

    So a2 = 2a1. Using vf2 = 2ad, I got vf2 / vf1 = √2(2a1)d / √2a1d

    So I got the second instance's velocity = √2(first instance's velocity)

    2) Conservation of energy

    U = Gm / r → U2 / U1 = ((4∏r)GAρ) / ((2∏r)GAρ)

    And so I got U2 = 2U1. Which means if the second instance has twice as much potential energy, it ends with twice as much kinetic energy. KE = mv2 / 2 → again I get √2 v

    Erhalkdjflnaeflkjafsfdlkji'msofrustrated

    Thanks in advance.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 25, 2014 #2

    Andrew Mason

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    If the force in the second case is always going to be double the force in the first, the change in kinetic energy will be double.

    ∫F2ds = ΔKE2 = ∫2F1ds = 2ΔKE1

    [i.e. ∫F1ds = mvf12/2

    ∫F2ds = mvf22/2

    Since mvf22/2 = ∫F2ds =
    ∫2F1ds = 2∫F1ds = 2mvf12/2 = m(√2vf1)2/2]


    ∴vf2 = √2vf1

    But the question is whether the premise is true for all distances.


    AM
     
    Last edited by a moderator: May 6, 2017
  4. Jan 26, 2014 #3

    Andrew Mason

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    The axial component of force on a unit mass located on the axis from an element of the ring, dm is:

    [tex]dF = \frac{Gdm}{(r^2 + R^2)}\sin{\beta}[/tex]

    where [itex]\sin{\beta}= \frac{r}{\sqrt{r^2+R^2}}[/itex] and r is the distance from the centre along the axis.

    So the total axial force is:

    [tex]\vec{F} = \int d\vec{F} = \frac{GM}{(r^2 + R^2)}\frac{r}{\sqrt{r^2+R^2}} = \frac{GMr}{(r^2 + R^2)^{3/2}}[/tex]

    The potential at a point on the axis a distance r from the centre is:

    [tex]U(r) = \int_0^r Fdr = \int_0^r \frac{GMr}{(r^2 + R^2)^{3/2}}dr[/tex]

    If you can solve that integral, you will have the answer.

    AM
     
  5. Jan 26, 2014 #4
    Hi Andrew Mason! :)

    I don't think it is necessary to do those integrals. Any point on the axis of ring is equidistant from every part of the ring so we can assert that the potential is simply:

    $$-\frac{GM}{\sqrt{z^2+R^2}}$$

    where ##z## is the distance of point from the centre of ring.
     
  6. Jan 26, 2014 #5

    Andrew Mason

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    Ok. So work out the change in potential from z>>R to z=0.

    What makes you think the correct answer is 2v?

    AM
     
    Last edited: Jan 26, 2014
  7. Jan 27, 2014 #6

    Andrew Mason

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    My last comment should have been directed at Agrasin: why do you think the correct answer is 2v? (hint: this is not the correct answer but neither is √2v).

    AM
     
  8. Jan 27, 2014 #7
    @Andrew Mason

    Sorry, I was looking at the wrong answer key. The correct answer is v, just v. Which is also confusing.

    Also, why are you using √z2+R2? If I'm not mistaken, this expression is the distance from the particle to any point along the ring. However, the ring is symmetrical and can just be taken as a point mass.

    In the second case, the point mass is twice the mass of the first point mass.
     
  9. Jan 27, 2014 #8

    Andrew Mason

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    The kinetic energy would double only if the force was always double (see post #2). But from the equation for force (post #3) you can see that this is not the case where z (i.e. r in my equation) is not >>R. When z = R the force in the second case is only 70% of the force in the first.


    It is easier to do the actual calculations using potentials. As Pranav-Arora pointed out, the potential is:

    [tex]U(z) = -\frac{GM}{\sqrt{(z^2 + R^2)}}[/tex]

    so for z = 0

    [tex]U(z) = -\frac{GM}{R}[/tex]

    The change in potential is equal and opposite to the change in KE per unit mass. Since the potential at z>>R is very small compared the potential at z = 0, the change in potential in going from z >> R to z=0 is essentially the potential at z=0.

    Work out those potentials in each of the two scenarios to compare the changes in kinetic energy per unit mass.

    AM
     
    Last edited: Jan 27, 2014
  10. Jan 27, 2014 #9
    Okay.

    U1 = (-GM) / R

    U2 = (-G(2M))/ R

    The second potential is twice the first. Therefore, the second change in kinetic energy is twice the first.

    Again, I'm arriving at v2 = √2v1
     
  11. Jan 27, 2014 #10

    AlephZero

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    The radius of the second ring is not R.
     
  12. Jan 27, 2014 #11
    But I thought the rings could be treated as point masses.

    And R in the gravitational potential equation is the distance from the particle (R, constant) to the point mass, I believe.
     
  13. Jan 27, 2014 #12

    lightgrav

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    no, because the Force (along the axis) actually goes to ZERO at the ring center ... there, all the individual Force contributions are _outward_

    (since acceleration is not constant, cannot use v²=2ad)
     
  14. Jan 27, 2014 #13
    So yes, it will accelerate less when it gets closer to the ring, where r<<R is no longer true. But the maximum velocity achieved by both particles can be calculated assuming the rings are point masses, can't it?
     
  15. Jan 27, 2014 #14

    Andrew Mason

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    No. The rings can be treated as approximately point masses only where z>>R. This is apparent from the expression for potential (ie. potential energy per unit mass).

    First of all you have to figure out where the maximum velocity will be. That occurs at z=?

    To find the change in kinetic energy (per unit mass), you just have to find the change in potential energy per unit mass (ie. the change in potential) between the starting point (where z>>R) and the point of maximum speed.

    As I explained, the potential at z>>R can be neglected as its absolute value is much, much smaller than the absolute value of the potential where z is close to R. As z increases the potential approaches 0.

    AM
     
  16. Jan 28, 2014 #15
    Maximum velocity occurs when R = 0, or when the particle goes through the center of the ring. Until then, acceleration is positive, but there, acceleration switches direction.

    So... The change in velocity is negligible until z = R ? If that's the case, I see how the velocities are equal because the change in potential energy (and kinetic energy) is equal in both instances between √(z2+r2) and z. But isn't z=R a pretty arbitrary choice? Couldn't you say the velocity is negligible only until 2z = R ?

    Thanks for your patience :-)
     
  17. Jan 28, 2014 #16

    Andrew Mason

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    R is a constant - it is the radius of the ring. Only z changes. Maximum velocity occurs when the rate of change of velocity i.e. acceleration (i.e. force) along the axis is 0, which occurs when z = 0 (r=0 in my #3 post).

    AM
     
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