Question

A 3.07 kg particle with velocity V= (7.02m/s)i - (7.14m/s)j is at X= 9.83m, Y= 2.36m. It is pulled by a 4.19N force in the negative x direction. About the origin, what are (a) the particle's angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is charging?

Answer 1

(a)

Position vector, r = 9.83 i^ + 2.36 j^

Linear momentum, p = mv = 3.07 (7.02m/s)i - (7.14m/s)j = 21.5514 i^ - 21.9198 j^

Angular momentum is cross product of position vector and linear momentum.

Angular momentum = r x p

= ( 9.83 i^ + 2.36 j^ ) x ( 21.5514 i^ - 21.9198 j^ ) =   -220.55 k^

Round off the result to 3 significant digits

Angular momentum = 221 kg.m²/s

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Net torque is cross product of the distance between point of application of force and point of observation and net force.

T = r x F = ( 9.83 i^ + 2.36 j^ ) x ( -4.19N i ) = ( 9.89 N.m ) k^

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(c) Rate of change of angular momentum is equal to net torque acting on the body ( calculated in part (b) )

Rate of change of angular momentum = ( 9.89 N.m ) k^