A 3.0 kg particle with velocity v = (5.0 m/s) ihat - (6.0 m/s)jhat is at x = 3.0

Question

A 3.0 kg particle with velocity v = (5.0 m/s) ihat - (6.0 m/s)jhat is at x = 3.0 m, y = 8.0 m. It is pulled by a 7.0 N force inthe negative x direction. (a) What is the angular momentum of the particle about theorigin? -174. kg·m2/s khat
(b) What torque about the origin acts on the particle? N·m khat
(c) At what rate is the angular momentum of the particlechanging? kg·m2/s2
solve this problem for me! Thanks A 3.0 kg particle with velocity v = (5.0 m/s) ihat - (6.0 m/s)jhat is at x = 3.0 m, y = 8.0 m. It is pulled by a 7.0 N force inthe negative x direction. (a) What is the angular momentum of the particle about theorigin? -174. kg·m2/s khat
(b) What torque about the origin acts on the particle? N·m khat
(c) At what rate is the angular momentum of the particlechanging? kg·m2/s2
solve this problem for me! Thanks

Explanation / Answer

Massm = 3 kg

Velocity v = 5i-6j

Force F = -7i

Position r =3i+8j

(b). torque T = rXF

                    = ( 3i+8j)X(-7i)

                    = (8*-7)(jXi)

                    = 56k   Since iXi = 0torque T = 56 Nm k

(c). rate is the angular momentum of the particle changing =magnitude of torque = 56 kg m ^ 2/ s^ 2

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