1) (10 pts)If you know the best time you ever had in a 100 m dash and you
   know your weight, estimate the momentum you had during that time. (If you
   do not know these values give your best guess.)

     I weigh 60 kg and in my students' time made 100m in 12s.
     The momentum p=m*v = 60 kg * 100m/12s = 500 kgm/s

2) (10 pts)It is possible to have a net torque when:
    a) all forces act through the axis of rotation
*   b) sum of all forces is equal to zero
    c) a body is in rotational equilibrium
    d) a body remains in unstable equilibrium


3) (10 pts) The moment of inertia of a rigid body
    a) depends on the axis of rotation
    b) cannot be zero
    c) depends on mass distribution
*   e) all of the above

4) (10 pts) If you are riding a bicicle forward, what is the direction of the
   angular momentum of the front wheel about its axle?

*   a) left
    b) right
    c) up
    d) down
    e) forward
    f) backward

5) (15 pts) An object at the surface of the Earth (at distance R from the
    center of the earth) weighs 90N. What is its weight at a distance 3R from
    the center?

       weight = mass * acceleration
       at distance R the acceleration is g= G*mass(Earth)/R^2
       at distance 3R the acceleration is G*mass(Earth)/(3R)^2
  --> the object weighs 1/9 the weight on the surface i.e., 10N

6) (10 pts) You are on a rotating amusement park ride shaped like a
   merry-go-round or a horizontal ferris wheel. You rotate with an angular
   velocity omega=1 rad/s. Suddenly without a change in the angular momentu L,
   the radius of the rotating wheel is halved. What is the magnitude of your
   new angular velocity?

       L = I * omega is conserved.
       I = mass * R^2
   when radius is halved I decreases by factor of 4. In order to keep L
   constant the angular velocity must increase by factor of 4.

7) (30 pts) A uniform plank is 12 m long and weighs 20kg. It is ballanced on a
   sawhorse at its center. An additional 40 kg weight is now placed on the left
   end of the plank. To keep the plank balanced the sawhorse has to be moved to
   what distance from the new weight along the plank?

       Balance requires the sum of all torques to be zero. 
       There are 3 torques: due to the additional weight, due to left side
       of the plank and due to the right side of the plank. 
       The sawhorse must be placed at a distance x from the left end. The
       length of the right side of the plank is y and the sum
         x + y = 12 m
       The weight of 1m of the plank is 20/12 kg
       The torque equation:
       40 kg * x/2 + 20/12*x * x/2 = 20/12*y * y/2
       x+y=12
       ---> the solution is x=2 m

8) (5 pts) A stick broken off a mahogany tree sinks in the sea water with
   density rho_salty but floats in a polluted water of a pond with density
   rho_dirty. What can you say about the relative denisties of water?

 * a) rho_dirty > rho_salty
   b) rho_dirty = rho_salty
   c) rho_dirty < rho_salty