- #1
JoeyBob
- 256
- 29
- Homework Statement
- See attached
- Relevant Equations
- L=r x p, Torque=change in L
First I calculated the momentum of m1. Since m2 was at rest after the collision, all its momentum was transferred, so m1 has a momentum of 158 i hat.
L=r x p, so its 916 k hat. This would also be the change in L because it was initially 0 when m1 had no velocity, so I know this is the net torque.
So I know the i hat component of torque is 0, so the normal force is the negative of the gravity force, so the friction force is
F=uFn=umg=-3.4496 i hat
Now Net toque=Torque from friction+Torque from Tension
Torque tension=916/(-5.8 x -3.4496)
=-45.782
Divide this by the radius and I get 7.89 N. The answer is suppose to be 3912.85 N
L=r x p, so its 916 k hat. This would also be the change in L because it was initially 0 when m1 had no velocity, so I know this is the net torque.
So I know the i hat component of torque is 0, so the normal force is the negative of the gravity force, so the friction force is
F=uFn=umg=-3.4496 i hat
Now Net toque=Torque from friction+Torque from Tension
Torque tension=916/(-5.8 x -3.4496)
=-45.782
Divide this by the radius and I get 7.89 N. The answer is suppose to be 3912.85 N