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Problems about torque and image charge

  1. Feb 10, 2013 #1
    http://farm9.staticflickr.com/8513/8460991004_d774e743cc.jpg

    1. As shown in the figure, a gyroscope consists of a uniform disk of radius r and an axle of length R through its center and along its axis. The other end of the axle is hinged on a table but its otherwise free to rotate in any direction. The gyroscope is spinning with angular velocity w with the axle inclined to the vertical direction. Find its angular velocity of precession.

    My attempt:
    What I thought is that the torque due to the weight of the gyroscope equals to its rate of change of angular momentum. mgRsin(theta) = d(Iw)/dt, and I get
    angular velocity = 2gR/(rw^2). Is it right?


    2. A point charge q is at x=3R/2 on the x-axis in front of a grounded conductor hemisphere of radius on a large conductor plate perpendicular to the x-axis and in the y-z plane. The center of the hemisphere is at (0,0,0). Find the potential energy energy of the point charge.

    My attempt:
    Can it be done using image charge? If so, how to set up the image charge configuration? Have no idea in how to deal with that hemisphere. If I put image charges to make the hemisphere zero potential, then the plane is not and I have to put another image charge to make the plane zero potential, but that will affect the potential of the hemisphere. So do I have to keep putting image charges (and end up with an infinite series)? Or are there other important things that I have omitted?

    thanks
     
    Last edited: Feb 10, 2013
  2. jcsd
  3. Feb 10, 2013 #2

    rude man

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    Gold Member

    Nope.

    Look at your dimensions - your angular velocity has dimensions of length ??? So you know that can't be right. Always check dimensions term-by-term. It's the most powerful checking tool there is.

    But you started right - τ = r x F with τ = dL/dt
    and |τ| = mgRsin(θ). τ = torque, L = angular momentum.

    Where you went wrong is understanding how angular momentum changes. You need to realize we are dealing with vectors here, and angular momentum is a vector that here changes not magnitrude but direction. In small interval of time Δt, ΔL = τΔt. You should draw this out & realize that ΔL is perpendicular to L. Just as dv/dt for circular motion is perpendicular to v - |v| stays constant but dv/dt changes direction and stays perpendicular to v.

    The last steps involve coming up with an expression for the precession angle Δψ which is the amount of precession in time Δt, in terms of τ, which you know, ΔL, L and θ (all scalars). You need to draw these vectors to understand how Δψ relates to L, ΔL, τ and θ. If you do it right you will be left with an expression for the precession frequency ωp = dψ/dt as a function of g, r, R, and w only. It might surprise you that it is not a function of θ.

    Get back to you on this one I hope.
     
  4. Feb 10, 2013 #3
    opps, just a typo, my answer to q1 is angular velocity = 2gR/(wr^2), is it correct (at least it has the dimension of 1/time)?
     
  5. Feb 10, 2013 #4

    rude man

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    Yes, that is correct. You must have been more closely on the right track than I gave you credit for! My apologies for my lengthy rambling ... :smile:
     
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