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?

a) The angular momentum of the particle is .

b) The torque acting on the particle is .

c) The rate at which the angular momentum is changing is .

a) The angular momentum of the particle (), in kilograms-meter per second, is given by the following vectorial formula:

(1)

Where:

If we know that , and , then the angular momentum of the particle is determined by determinants:

The angular momentum of the particle is .

b) The torque (), in newton-meters, acting on the particle is defined by the following expression:

(2)

Where is the force, in newtons, acting on the particle.

If we know that and , then the torque experimented by the particle is determined by determinants:

The torque acting on the particle is .

c) The rate at which the angular momentum changes in time is found deriving (1):

(3)

In other words, the rate of a constant mass system at which the angular momentum changes is equal to the torque experimented by the particle. Hence, the rate at which the angular momentum changes is  .

The rate at which the angular momentum is changing is .

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Explanation:

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 = <(3.0 m) i + (8.0 m) j> × <(15 kg m/s) i + (-18 kg m/s) j>

L = (-174 kg m²/s) k

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

τ = r × F

τ = <(3.0 m) i + (8.0 m) j> × <(-7.0 N) i + (0 N) j>

τ = (56 Nm) k

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

L' = τ

L' = (56 Nm) k