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Conservation of Energy and Gravitation

  1. Nov 22, 2007 #1
    1. A block slides along a track with elevated ends. The flat part has length L = 0.2m, and the object is released from a height of 0.1m. The curved portion of the track is frictionless, but the flat part has uk = 0.15 . where does the object finally come to rest?

    Etotal = Ui + Ki = Uf + Kf + work done by friction
    mgy + 0 = mgy' + 0 + ukFn
    mg(.1) = mgy' + (.15)mg
    .1 = y' + .15
    y' = -.05

    this doesnt seem right, will I have to do multiple calculations of conservation of energy, since there is a section with friction and a section without. Do I have to consider angular momentum since its sliding down a slope at first?

    2. Several planets possess nearly circular surrounding rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous ring of mass M and radius R.
    a) Find an expression for the gravitational force exerted by the ring on a particle of mass m located a distance x from the center of the ring along its axis.
    b) Suppose that the particle falls from rest as a result of the attraction of the ring of matter. Find an expression for the speed with which is passes through the center of the ring.

    since we are dealing with a ring, should I use inertia of the ring instead of mass?

    F = -GMm/r^2
    I of ring = mr^2

    F = -G(MR^2)m/x^2

    W = Fd = -G(MR^2)m/x

    Etotal = U - K
    -G(MR^2)m/x = (1/2)mv^2
    v = -sqrt(2G(MR^2)/x)

  2. jcsd
  3. Nov 22, 2007 #2

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    1.The initial KE at the start of the flat portion is equal to the energy gained after dropping through a height of 0.1 m. This KE will be equal to work done by friction in bringing it to rest. Equate the two.

    2. a) This has got nothing to do with MI. All the points on the ring is at an equal dist from any pt on the axis. You know the dist. Just sum the forces. Only the components of the forces along the axis will remain. The components perp to the axis will cancel out for diametrically opp points.

    2. b) Can you find the potential at any point along the axis? Remember, force = -grad phi. After that, final E = initial E, because gravitation is a conservative force.
    Last edited: Nov 22, 2007
  4. Nov 22, 2007 #3
    mgh = kf = work done by friction to bring it to rest
    mgh = fd
    mgh = uk*n*d
    mgh = uk*mg*d
    h = uk*d
    d = h/uk
    d = .67

    but this is greater than the length, so will the brick go upt he other side?
    would i have to do this a few times and add the energies lost each time?

    a) -GMm/x^2
    is this all I need to leave it at?

    U = mgx
    mgx = 0 when x = 0
    so when U = 0
    K = mgx
    mv^2 = mgx
    v = sqrt(gx)

    is this correct?

  5. Nov 23, 2007 #4

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    (Hi, posting problems separately makes it easier for all to share in the helping.)

    1. Right. Since the elevated ends will not take away energy, it seems the mass will traverse the flat portion three times and after that will come to rest 0.07 m from the
    other end than from which it started.

    2a. No, you have to sum it and express it in terms of x, M and R. Draw a diagram. Remember that the dist d of the pt P on the axis from a pt on the ring is given by
    d^2=R^2+x^2. Take component of the force at P along and perp to axis.

    2b. No, the U you've written is for uniform g field. You have to use the force you get from 2a and integrate. Do 2a first.

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