A uniform disk of mass m = 3.4 kg and radius R = 5 cm can rotate about an axle t

Question

A uniform disk of mass m = 3.4 kg and radius R = 5 cm can rotate about an axle through its center. Four forces arc acting on it as shown in the figure. Their magnitudes are F_1 = 1.5 N, F_2 = 1.5 N, F_3 = 8.5 N and F_4 = 7.5 N. F_2 and F_4 act a distance d = 15 cm from the center of mass. These forces are all in the plane of the disk. Randomized Variables m = 3.4 kg R = 5 cm F_1 = 1.5 N F_2 = 1.5 N F_3 = 8.5 N F_4 = 7.5 N d = 1.5 cm Write an expression for the magnitude tau _1 of the torque due to force F_1. Calculate the magnitude tau _1 of the torque due to force F_1, in N middot m. Write an expression for the magnitude tau _2 of the torque due to force F_2. Calculate the magnitude tau _2 of the torque due to force F_2, in N middot m. Write an expression for rite magnitude tau _3 of the torque due to force F_3. Write an expression for the magnitude tau _4 of the torque due to force F_4. Calculate the magnitude tau _4 of the torque due to force F_4, in N middot m. Calculate the angular acceleration alpha of the disk about its center of mass in rad/s^2. Let the counter-clockwise direction be positive. alpha = _______

Explanation / Answer

(A) torque1 = R F1

(B) torque1 = (0.05 m ) (1.5) = 0.075 N m

(C) torque2 = d F2 sin45

(D) torque2 = (0.015) (1.5) (sin45)

= 0.016 N m

(E) torque3 = R F3 sin0 = 0

(f) torque4 = d F2 sin53

(g) torque4 = 0.015 x 7.5 x sin53 = 0.09 N m

(h) Net torque = 0.075 - 0.016 + 0.09 = 0.1488 N m

I = 3.4 x 0.05^2 /2 = 4.25 x 10^-3 kg m^2

Net torque = I alpha

alpha = 35 rad/s^2

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