Question

At the instant of the figure, a 4.10 kg particle P has a position vector r of magnitude 9.10 m and angle θ1 =  44.0° and a velocity vector v of magnitude 4.60 m/s and angle θ2= 33.0°.  Force F,of magnitude 5.30 N and angle θ3 = 33.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle? 

Answer 1

a)

angular momentum = r X mv

= r x mv x sinθ

= 9.10 x 4.1x 4.60 x sin(180 - ₂)

= 171.626 sin (180 - 33 )

= 93.47 kg m² / s

b)

Torque = r X F

= r x F x sinθ

= 9.10 x 5.30 x sin30⁰

= 24.11 N-m