At the instant of the figure, a 1.3 kg particle P has a position vector r vector

Question

At the instant of the figure, a 1.3 kg particle P has a position vector r vector of magnitude 1.5 m and angle theta_1 = 45 degree and a velocity vector V vector of magnitude 4.1 m/s and angle theta_2 = 30 degree Force F vector of magnitude 1.4 N and angle theta_3 = 30 degree acts on P. All three vectors lie in the xy plane. (Express your answers in vector form.) (a) What is the angular momentum of the particle about the origin? L vector = kg m^2/s (b) What is the torque acting on the particle about the origin? T vector = N middotmiddotm

Explanation / Answer

m = 1.3 kg ; r = 1.5 m ; theta-1 = 45 deg ; v = 4.1 m/s

theta-2 = 30 deg ; F = 1.4 N ; theta-3 = 30 deg

a)We know that, angular momentum is :

L = m v r sin(theta)

L = 1.3 x 4.1 x 1.5 x sin30 = 3.998 kg-m^2/s

Hence, L = 3.998 = 4 kg-m^2/s

b)We know that torque is given by:

Tau = r F sin(theta)

Tau = r F sin(theta-3) = 1.5 x 1.4 x sin30 = 1.05 N-m

Hence, Tau = 1.05 N-m

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