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makkiz [27]
3 years ago
8

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?
Physics
1 answer:
Step2247 [10]3 years ago
6 0

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

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