A 6.97 kg particle moves in an xy plane. At the instant when the particle\'s pos

Question

A 6.97 kg particle moves in an xy plane. At the instant when the particle's position and velocity are r =(5.00i+5.00j)m , the force on the particle is At this instant, determine (a) the particle's angular momentum about the origin, (b) the particle's angular momentum about the point x = 0, y = 5.00 m, (c) the torque acting on the particle about the origin, and (d)the torque acting on the particle about the point x = 0, y = 5.00 m. Assume counterclockwise direction positive and clockwise direction negative.

Explanation / Answer

a)6.97*7.68*5

=267.648 kg m^2/s

b)torque= -(5*7.38)

=-36.9 clockwise

c)-36.9

clockwise

d)-36.9

clockwise

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