Need Help With This Physics Problem

Question

A set of four masses is formed into a square of side l using four wires that have negligible mass, as shown, by astronauts who took physics, are bored, and need to pass the time in the ISS. They amuse themselves by trying to predict the motion of weird objects like this in the apparent zero gravity (remember gravity acts on everything in the ISS and on the ISS itself, but all objects are in free fall toward the center of Earth at the same rate as the ISS and so appear to have no gravity on them!) inside the ISS.

**You can pick and choose questions that you would like to answer. Any help is appreciated:

A force F is applied for a short time Delta t to the object, which is stationary (with respect to the ISS) by an astronaut's finger, in the direction shown.

E) What is the torque (give a magnitude and direction) on the object during the time that the

force is applied?

(F) What is the change in the angular momentum of the object resulting from the application of the

force?

(G) What is the angular velocity of the object (give a magnitude and direction) after the force is applied?

(H) What is the kinetic energy of the object?

http://i36.photobucket.com/albums/e12/lilbac2003/BoredAstronauts.gif

Answer 1

E)   Torque = radius x force

       given that the effective radius is a line perpendicular line form the center of mass to the side the force is being applied to, and that the force is in a direct line with that side IE tangential   the radius is 1/2 unit

torque = 1/2  times force.

F) torque = change in angular momentum / time

     change in angular momentum = torque x time

     change in angular momentum  = 1/2 (force)(time)

g)  I always have trouble with this I hope the link helps

H  the  kenetic energy will  be equal to force x time

    It will find expression as 6m V but the point is it will equal the energy put in the system by the force x time the force is applied.

Answer 2

E)  T = F*l/2

F)  L = T*dt

G)  w = L/I where I is the moment of inertia about the CM = 3*l*√2*m where l is the length of 1 side.

H)  KE = KEl + KEr = ½Mv² + ½Iw² where M = 6m, v = Fdt/6m