A uniform thin rod of mass m = 4.5 kg and length L = 1.5 m can rotate about an a

Question

A uniform thin rod of mass m = 4.5 kg and length L = 1.5 m can rotate about an axle through its center. Four forces arc acting on it as shown in the figure. Their magnitudes arc F_1 = 8.5 N, F_2 = 1.5 N, F_3 = 12.5 N and F_4 = 19.5 N. F2 acts a distance d = 0.15 m from the center of mass. Calculate the magnitude tau_1 of the torque due to force F_1 in N-m. Calculate the magnitude tau_2 of the torque due to forece F_2 in N-m. Calculate the magnitude tau_3 of the torque due to force F_3, in N-m. Calculate the magnitude tau_4 ol the torque due to force F_4 in N-m. Calculate the angular acceleration alpha of the thin rod about its center of mass in rad/s2. Let the counter clockwise direction be positive.

Explanation / Answer

Torque on an object is given by= r (cross) F

a) Torque due to F1 = -0.75 * 8.5 = -6.375 Nm

b) Torque due to F2 = d*F2*sin(45)= 0.15 * 1.5* 1/1.414 = 0.159 Nm

c)Torque due to F3 = 0 Nm (because r=0)

d) Torque due to F4 =r*F4*sin(0) = 0 Nm

e) As, net Torque= I*(alpha)

-6.375+0.159 = 4.5 * 1.52* alpha/ 12

alpha = -7.36 rad/sec2

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