At the instant of the figure, a 6.60 kg particle P has a position vector of magn

Question

At the instant of the figure, a 6.60 kg particle P has a position vector of magnitude 4.90 m and angle ?1 = 48.0 and a velocity vector of magnitude 8.90 m/s and angle ?2 = 32.0. Force , of magnitude 2.40 N and angle ?3 = 30.0 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?

At the instant of the figure, a 6.60 kg particle P has a position vector of magnitude 4.90 m and angle ?1 = 48.0½ and a velocity vector of magnitude 8.90 m/s and angle ?2 = 32.0½. Force , of magnitude 2.40 N and angle ?3 = 30.0½ 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?

Explanation / Answer

Angular Momentum = r x mV = m*r*v*sin (180 - 2) = 1525.245 kg·m2/s

Torque = r x F = F*r*sin 3 = 5.88 N m

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