Problem Set 1 solutions

1.  Setting the positive y-axis up and the positive x-axis to the right, we see from the figure that as the ball bounces on the floor, it experiences a tangential force F_x that changes the ball's x-momentum:

int(F_x dt) = Dp_x = m(v_xf - v_xi)

where int stands for the integral, p_x is the x-component of the momentum, m is the ball's mass, v_xf is the final x-component of the velocity, and v_xi is the initial x-component of the velocity.  The tangential force also produces a torque on the ball:

torque = int(aF_xz dt) = DL = I(w_f - w_i)z

where z is the unit vector in the positive z-direction (out of the page), L is the angular momentum, w_f is the final angular speed, w_i is the initial angular speed, and I is the ball's rotational moment of inertia 2ma²/5.  Putting the first equation into the second for the integral gives:

v_xf - v_xi = 2a(w_f - w_i>)/5

For the ball to bounce back and forth as pictured (i.e. symmetrically), we must have v_xf = -v_xi and w_f = -w_i.  Substituting these into the above equation gives the condition for the initial throw:

v_xi = 2aw_i/5

2.  For this problem, I will set the positive x-axis in the direction of motion and the origin at the edge of the dust cloud.  Since there are no external forces involved, momentum must be conserved.  So, the spacecraft's original momentum, Mv₀, will be equal to the momentum of the craft (including the dust) at all times.  The momentum at any time will be the ship's speed multiplied by its total mass (M plus the mass of all the dust that is stuck to it at that point):

p = Mv₀ = (M+Axr)v = constant

But v is just dx/dt.  Putting that into the above equation and distributing it into the parentheses gives:

Mv₀ = M dx/dt + Ar xdx/dt

The right-most term can be rewritten as:

Ar xdx/dt = Ar/2 d/dt (x²)

Putting that back into the differential equation just above the one above and factoring out the differential operator gives:

Mv₀ = d/dt (Mx + Ar x²/2)

Integrating both sides with respect to t gives:

Mv₀ t = Mx + Ar x²/2 + c

With the initial condition that at t=0, x=0, the integration constant c must be 0 as well.  Now, we need to solve for x.  The equation above is just a standard quadratic that can be solved using the quadratic formula.  Doing so and keeping the solution that yields increasing x for increasing t gives:

x(t) = -M/Ar  [1 - sqrt(1+2Av₀r t/M)],  for t>0

3.  Well, in order for the skyhook satellite to remain in orbit, the total gravitational force on the rope must equal its centripetal "force":

int(GMl/r² dr) evaluated from R to (R+ L) = int(rw²l dr) evaluated from R to (R+L),

where G is the universal gravitational constant, M is the mass of the Earth, l is the (constant) linear mass density of the rope, R is the radius of the Earth, w is the (constant) rotational speed of the Earth, and L is the (unknown) length of the rope (we use the Earth's rotational speed because we know that the rope is in geosynchronous orbit; the rope always appears suspended above the same point on the Earth so it must have the same rotational speed).  Pulling out and canceling the common l factor and then performing the integration and evaluations gives:

GML/{R(R+L)}= w²(L²+2RL)/2

We want to find L, so we need to solve this quadratic equation for L.  After all the algebra, we find

L = [-3R + sqrt{9R²+4(2GM/Rw² - 2R²)}]/2

With R=6.4X10⁶m, GM/R²=g=9.8m/s², and w=2prad/24hours=7.3X10^-5rad/s, L=1.5X10⁸m.

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