A uniform disk of mass m = 1.2 kg and radius R = 11 cm can rotate about an axle

Question

A uniform disk of mass m = 1.2 kg and radius R = 11 cm can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F_1 = 6.5 N, F_2 = 1.5 N, F_3 = 7.5 N and F_4 = 3.5 N. F_2 and F_4 act a distance d = 1.5 cm from the center of mass. These forces are all in the plane of the disk. Randomized Variables m = 1.2 kg R = 11 cm F_1 = 6.5 N F_2 = 1.5 N F_3 = 7.5 N F_4 = 3.5 N d = 1.5 cm Write an expression for the magnitude tau_1 of the torque due to force F_1. tau_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 of force F_2, in N middot m. Write an expression for the 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.

Explanation / Answer

(A) torque1 = R F1

(B) torque1 = 0.11 x 6.5 = 0.715 N m

(C) torque2 = - d F2

(D) torque2 = 0.015 x 1.5 = - 0.0225 N m

(E) torque3 = R F3 sin0 = 0

(F) torque4 = d F4 sin53

(g) torque4 = 0.015 x 3.5 x sin53 = 0.0419 N m

(H) net torque = 0.715 - 0.0225 + 0.0419 = 0.734 N m

I = m r^2 /2 = 1.2 x 0.11^2 / 2 = 7.26 x 10^-3

net torque = I alpha

alpha = 0.734 / (7.26 x 10^-3)

= 101.1 rad/s^2

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