A uniform thin rod of mass m = 1.5 kg and length L = 1.8 m can rotate about an a

Question

A uniform thin rod of mass m = 1.5 kg and length L = 1.8 m can rotate about an axle through its center. Four forces are acting on it as shown in the figure. Their magnitudes are F1 = 7 N, F2 = 3 N, F3 = 12 N and F4 = 15.5 N. F2 acts a distance d = 0.13 m from the center of mass.

Randomized Variablesm = 1.5 kg
L = 1.8 m
F1 = 7 N
F2 = 3 N
F3 = 12 N
F4 = 15.5 N
d = 0.13 m 20% Part (a) Calculate the magnitude ?1 of the torque due to force F1 in N?m.
20% Part (b) Calculate the magnitude ?2 of the torque due to force F2 in N?m. 20% Part (c) Calculate the magnitude ?3 of the torque due to force F3 in N?m. 20% Part (d) Calculate the magnitude ?4 of the torque due to force F4 in N?m.
20% Part (e) Calculate the angular acceleration ? of the thin rod about its center of mass in rad/s2. Let the counter-clockwise direction be positive.

Explanation / Answer

torque = force X distance

torque due to F1

Tf1 = 7 * 0.9 * sin(90)

Tf1 = 6.3 N-m

torque due to F1 = 6.3 N-m

torque due to F2

Tf2 = 3 * 0.13 * sin(45)

Tf2 = 0.2757 N-m

torque due to F2 = 0.2757 N-m

torque due to F3

Tf3 = 12 * 0 * sin(60)

Tf3 = 0 N-m

torque due to F3 = 0 N-m

torque due to F4

Tf4 = 15.5 * 0.9 * sin(0)

Tf4 = 0 N-m

torque due to F4 = 0 N-m

moment of inertia of rod about its center = 1/12 * M * L^2

moment of inertia of rod about its center = 1/12 * 1.5 * 1.8^2

moment of inertia of rod about its center = 0.405

total torque = Tf1 - Tf2

total torque = 6.3 - 0.2757

total torque = 6.0243

total torque = moment of inertia * angular acceleration

6.0243 = 0.405 * angular acceleration

angular acceleration = -13.3873 rad/sec^2

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