at time t = 0, a 3.0 kg particle with velocity v = (5.0 m/s) i - (6.0 m/s) j is at x = 3.0 m, y = 8.0m. it is pulled by a 7.0 N force in the negative x direction. About the origin, what are the particle's angular momentum, the torque acting on the particle and the rate at which the angular momentum is changing?

Question

Winston Lucero

Physics

6 January, 11:07

at time t = 0, a 3.0 kg particle with velocity v = (5.0 m/s) i - (6.0 m/s) j is at x = 3.0 m, y = 8.0m. it is pulled by a 7.0 N force in the negative x direction. About the origin, what are the particle's angular momentum, the torque acting on the particle and the rate at which the angular momentum is changing?

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Jayvion · Answer 1

The radius, velocity, and force vectors are:

r = (3.0 m) i + (8.0 m) j

v = (5.0 m/s) i - (6.0 m/s) j

F = (-7.0 N) i

Angular momentum is the cross product of the radius vector and the linear momentum vector:

L = r * p

L = r * (mv)

L = *

L = (-174 kg m²/s) k

Torque is the cross product of the radius vector and the force vector:

τ = r * F

τ = *

τ = (56 Nm) k

Rate of change of angular momentum is equal to the torque.

L' = τ

L' = (56 Nm) k