Practice Solutions

Physics 5200

  1. (10 points) A rubber ball is dropped from a height of 5m. When it strikes the ground, it rebounds with 80% of its incident speed. What height does the ball reach after 1 rebound?

    Use conservation of energy. When the ball hits the ground its kinetic energy will equal its starting potential energy:

    1/2 mv0² = mgh0

    thus

    v0 = (2gh0)½.

    On rebound its speed will be v1 = 0.8v0. It will rise to a height h1 where its kinetic energy is converted to potential energy:

    1/2 mv12 = mgh1

    thus

    h1 = v12/2g = 0.64h0.

  2. (10 points) An object of mass 2kg moves in the x-y plane, acted on by a force
    F = ((3/x²) - 12x²y³)i + ((2/y³) - 12x³y²)j
    1. Is this force conservative?
      The force is conservative if its curl is zero.
      ∇×F = -36x²y² + 36x²y² = 0, so yes, F is conservative.
    2. If the object has a speed of 5m/s when it is located at r = i + j what is its speed at the location r = 3i + ½ j?
      Since F is conservative, we can make use of conservation of energy, namely, T0 + V0 = T1 + V1 where 0 is the initial location and 1 is the final location. By inspection we can determine that V = (3/x) + (1/y²) + 4x³y³. Therefore, V0 = 3+1+4 = 8J and V1 = 1+4+27/2 = 18.5J. T0 = 1/2mv2 = 25J, so it must be that T1 = 14.5J or v1 = (2*14.5/2)½ = 3.8m/s.
  3. (10 points) A star of mass M = 2×1030kg and radius R0 = 7×108m rotates with angular velocity ω0. Suddenly the star collapses under the force of its own gravity and its radius is reduced to R1 = 104m. (For this problem, collapsed means that the radius is reduced while the mass remains constant. Assume that the star remains spherical.)
    1. What is the angular velocity of the collapsed star? For a uniform sphere, the moment of inertia about an axis through the center is Icm = (2/5)MR².
      There are no external forces on the star, therefore angular momentum is conserved.
      L = I0ω0 = (2/5)MR0²ω0 = I1ω1 = (2/5)MR1²ω1
      Therefore ω1 = ω0(R0²/R1²) = 5×108ω0 .
    2. Stability of the collapsed star demands that the gravitational force at the surface exceeds the centrifugal force. Use this fact to derive a limit on the angular velocities after collapse, ω1, and before collapse, ω0.

      F/dm = GM/R1² > ω1²R1 
      and therefore, ω1 < (GM/R1³)½ = 11500/sec.
      Note that the velocity of a point on the surface of the collapsed star is 1.1×108m/s, more than ⅓ the speed of light. And to reach this rate of rotation, the uncollapsed star need rotate at a rate of ω0 = 2×10-5/sec, or only f=3×10-6Hz, that is 10 rotations per year.
  4. (10 points) If the motion of a one dimensional oscillator is not sinusoidal, is the oscillator harmonic (i.e. is the motion governed by an equation of the form mx[ddot] + kx = 0)?

    Absolutely not. The general solution for any harmonic oscillator is a sinusoidal function. The only way to get non-sinusoidal motion is to not have an harmonic oscillator.

  5. (a) (5 points) If a simple harmonic oscillator has energy E and amplitude A, what's the value of the spring constant k?

    Energy is the sum of kinetic plus potential energies for a simple harmonic oscillator. At a maximum of its displacement, the kinetic energy of the oscillator is zero, so all of its energy is potential energy: E = (1/2)kA2, from which we find k = 2E/A2.

    (b) (5 points) If this oscillator has a damped frequency ωd, and it takes n complete oscillations for the amplitude to decrease to 1/e of its initial value, what is the mass of the oscillator.

    e-nβTd = e-1 therefore, β = 1/nTd = ωd/2nπ. But ω0² = ωd² + β² = ωd²(1 + 1/4n²π²) = k/m. Solving for m yields:
    m = 2E(Aωd)-2(1 + 1/4n²π²)-1.
    This result is expressed in terms of known quantities.

  6. (10 points) An electron moves in a region of space with a constant magnetic field, B = Bk, and a constant electric field, E = Ej. Write down the x, y, and z equations of motion for the electron.

    For a charge q moving in electric and magnetic fields, the Lorentz force is given by F = q(E + v×B). For the given fields this becomes:
    F = q(Ej + y[dot]Bi - x[dot]Bj) = mr[ddot]   so the components are
    mx[ddot] = y[dot]B
    my[ddot] = qE - x[dot]B
    mz[ddot] = 0

  7. (10 points) A particle with mass m moves in one dimension subject to a force F = -kx - 2(mk)1/2x[dot].

    1. Set up the equation of motion and show that
      x = Ate-βt + Be-βt 
      is a solution, where A and B are constants of integration and β² = k/m.

      The equation of motion can be written:
      mx[ddot] + 2(mk)½x[dot] + kx = 0.
      The derivatives of the proposed solution are:
      x[dot] = -βAte-βt + (A - βB)e-βt  and  x[ddot] = β²Ate-βt - β(2A - βB)e-βt
      Upon inserting these expressions into the equation of motion, and canceling out the common factor of e-βt, we have:
      At{mβ² -2β(mk)½ + k} + mβ(βB - 2A) + 2(mk)½(A - βB) + kB = 0
      The expression in curly braces, multiplied by t must be zero independent of the remaining part. That expression gives us β² = k/m. Then the remaining expression becomes:
      kB -2kB + kB -2(mk)½A +2(mk)1/2A = 0.
      This is obviously satisfied for any values of A and B, so we have shown that the proposed solution works. Additionally, since this solution contains two independent constants of integration for a second order differential equation, it must be a complete solution.

    2. Given that β = 1sec-1 and x[dot](t=1sec) = -0.736m/s and x(t=1sec) = 0.736m, determine the values of the constants of integration A and B.
      x(t=1sec) = (A+B)e-1 = 0.736m  and   x[dot](t=1sec) = -Be-1 = -0.736m. Therefore, B = 2.00m and A = 0m/s.

  8. (10 points each) A truck at rest has its rear door fully open. The truck accelerates forward at constant rate A, and the door begins to swing shut. The door is uniform, with mass M, height h, and width w, and negligible thickness. Neglect air resistance.

    1. Where is the center of mass of the door, and what is its moment of intertia about the hinge (the hinge is along one edge of the door)?
      The center of mass of the door is at its geometric center, h/2 from the top or bottom edge, and w/2 from the left or right edge. The moment of inertia is, using the parallel axis theorem, (1/12)Mw² + M(w/2)² = (1/3)Mw².

    2. Find the instantaneous angular velocity of the door about its hinge when it has swung through 90°.
      The truck is uniformly accelerating horizontally, so in a frame moving with the truck, it appears as though there is a horizontal component to gravity with a value of -A. Conservation of energy then gives: MAw/2 = 1/2 Iω² or ω = (3A/w)½.

  9. (10 points) A small block of mass m sits on a disk rotating with angular velocity ω. The coefficient of friction between the block and the disk is μ. At an instant of time the block is located a distance x0 from the center of rotation, with an arbitrary velocity. Write down all the apparent forces (including fictitous forces) acting on the block as seen from a reference frame rotating with the disk.
    Place the origin or coordinates at the center of rotation, and fix the x axis to point to the block. Then the coordinate system rotates with the angular velocity of the disk, ω = &omega k. The block is located at r = x0i, and let's assume it has velocity v = vxi + vyj. The forces acting on the block are:
    Centrifugal force: Fcentf = mω²x0i 
    Coriolis force: Fcor = -2mω×v = 2mω(vyi - vxj)
    Friction: Ff = -μmgv/|v|.

  10. (10 points) A bead slides along a circular loop of wire without friction. The loop sits vertically and rotates about an exis passing through its center, with constant angular velocity ω. The loop has radius a, and gravity points downward. The angle from the axis of rotation to the bead is . There are 3 angles for which the bead is in equilibrium, that is, F = 0.

    1. Set up the equations of motion in an appropriate coordinate system, and find the 3 equilibrium angles.
      Let's choose a coordinate system fixed to the bead. This will then look similar to the Earth based problems we worked. In this system, ω = ωsinλ j + ωcosλ k. (The difference from earlier problems is due to the definition of λ.) A0 points towards the axis of rotation and is given by A0 = ωa²sinλ(cosλ j - sin&lambda k). For the bead, r = 0, and v = vj. So, the transverse and centrifugal forces are zero, and the coriolis force is Fc = -2mω×v = -2mvωa²sin²&lambda i. The remaining forces are the force of gravity, and the force of the wire on the bead. Fg = -mg(sinλ j + cosλ k). The x and z components of the forces must be canceled by the force the wire exerts on the bead. The resulting force is:
      Fy = -mωa²sinλ cos&lambda - mgsinλ = -msinλ(ωa²cosλ + g).
      This force is zero when sinλ=0 at λ=0° or 180°, or when λ = cos-1(-g/ωa²).

    2. If the loop rotates too slowly, one of the equilibrium angles disappears (two of the angles merge into one). What is the rotation rate where this occurs?
      If the loop rotates too slowly such that g/ωa²>1, then the third angle listed above merges with the 180° solution. This occurs for ω<g/a².