Torque:

$\overline{){\mathbf{\tau}}{\mathbf{=}}{\mathbf{r}}{\mathbf{\xb7}}{\mathbf{F}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{\mathbf{\theta}}}$

Moment of inertia of a rod about its center of mass:

$\overline{){\mathbf{I}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{3}}{\mathbf{M}}{{\mathbf{L}}}^{{\mathbf{2}}}}$

Angular acceleration and torque are related by:

$\overline{){{\mathbf{\tau}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}{\mathbf{=}}{\mathbf{I}}{\mathbf{\alpha}}}$

Where I is the moment of inertia.

**(A)**

F = 1.5 N

r = L/2 = 1.2/2 = 0.6 m

θ = 90°

A uniform thin rod of mass m = 2.6 kg and length L = 1.2 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} = 1.5 N, F_{2} = 3.5 N, F_{3} = 11 N, and F_{4} = 15.5 N, F_{2} acts as a distance d = 0.28 m from the center of mass.

Part A. Calculate the magnitude τ_{1} of the torque due to force F_{1} in N•m.

Part B. Calculate the magnitude τ_{2} of the torque due to force F_{2} in N•m.

Part C. Calculate the magnitude τ_{3} of the torque due to force F_{3} in N•m.

Part D. Calculate the magnitude τ_{4} of the torque due to force F_{4} in N•m.

Part E. Calculate the angular acceleration a of the thin rod about its center of mass in rad/s^{2}. Let the counter-clockwise direction be positive.

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