A uniform thin rod of mass m = 3.2 kg and length L = 1.5 m can rotate about an a

Question

A uniform thin rod of mass m = 3.2 kg and length L = 1.5 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F_1 = 2.5 N, F_2 = 3.5 N, F_3 = 15 N and F_4 = 17 N. F_2 acts a distance d = 0.16 m from the center of mass. Calculate the magnitude tau_1 of the torque due to force F_1, in newton meters. tau_1 = Calculate the magnitude tau_2 of the torque due to force F_2 in newton meters. Calculate the magnitude tau_3 of the torque due to force F_3 in newton meters. Calculate the magnitude tau_4 of the torque due to force F_4 in newton meters. Calculate the angular acceleration alpha of the thin rod about its center of mass in radians per square second. Let the counterclockwise direction be positive.

Explanation / Answer

(A) torque1 = - (L/2) (F1)

= - (1.5/2) (2.5)

= - 1.875 N m

(B) torque2 = d F2 sin45 = 0.16 x 3.5 x sin45

= 0.396 N m

(C) torqe3 = 0 { r = 0 }

(D) torque4 = (L/2) F4 sin0 = 0

(E) net torque = - 1.875 + 0.396 = - 1.479 N m

moment of inertia = m L^2 / 12 = 3.2 x 1.5^2 / 12

= 0.6 kg m^2

net torque = I alpha

alpha = - 1.479 / 0.6 = - 2.46 rad/s^2

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