Question

6. Angular Momentum and Torque on a Meteor A meteor enters Earth's atmosphere and is observed by someone on the ground before it burns up in the atmosphere. The vector 25 m i +25 m j gives the position of the meteor with respect to observer. At the instant the observer sees the meteor, it has linear momentum p 1.5 kg -2.0) m/s 3 and it is accelerating at a constant rate of 2.0 m/s(-3) along its path, which for our purposes can be taken as a straight line. Given (nn = 1.5 kg, azー0, a,--2.0m/s2Nzー0,v,--2-(2.0)t). (a) What is the angular momentum of the meteor about the origin, which is the location of the observer? (1 points) Find the cross product product between F and p) (b) Take the time derivative of the angular momentum. (1.0 points) c) Find the angular momentum at time t 0 (1.0 points) (d) What is the force here provided that you know mass and acceleration above (1.0 points) (e) What is the net torque in the system? (1.0 points)

Answer 1

1.5 (-2)クm15 -2 C-2) t Ma