A uniform thin rod of mass m = 0.75 kg and length L = 1.5 m can rotate about an

Question

A uniform thin rod of mass m = 0.75 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 = 95 N, F_2 = 35 N, F_3 = 15 N and F_4 = 16 N. F_2 acts a distance d = 0.13 m from the center of mass. (a) Calculate the magnitude tau_1 of the torque due to force F_1, in newton meters. (b) Calculate the magnitude tau_1 of the torque due to force F_2 in newton meters. (c) Calculate the magnitude tau_3 of the torque due to force F_3 in newton meters. (d) Calculate the magnitude tau_4 of the torque due to force F_4 in newton meters. (e) 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. alpha =

Explanation / Answer

a)

here

torque t = r * F * cos(theta)

So

t1 = 9.5N * 0.75m * cos0 = 7.1 Nm

b)

t2 = 3.5N * 0.13m * cos45 = 0.3217 Nm

c)

t3 = 15 N * 0m * cos30 = 0 Nm

d)

t4 = 16N * 0m * cos90 = 0 Nm

e)

Total torque, where counterclockwise is positive, is

T = T2 - T1 = 0.3217 - 7.1 = -6.7783 Nm

The moment of inertia of a rod about its center is

I = m * L^2 / 12

I = 0.75 * (1.5)^2 / 12 = 0.140625 kg·m²

T = I * alpha

-6.7783 = 0.140625 * alpha

alpha = -48.2 rad/s^2

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