The angular velocity of the uniform 50-lb slender rod when rotated 90 degrees is determined as 4.34 rad/s.
The net torque on the slender rod is calculated as follows;
∑τ = 0
Iα = Frsinθ
Iα = τsinθ
where;
[tex]\alpha = \frac{\tau \times sin(90)}{I} = \frac{\tau \times sin(90)}{\frac{1}{12}ML^2 }[/tex]
L = (100. lb.ft)/(50 lb) = 2 ft
[tex]\alpha = \frac{100 \times sin(90)}{\frac{1}{12} (50 \times 2^2)} \\\\\alpha = 6 \ rad/s^2[/tex]
[tex]\omega _f^2 = \omega_i^2 + 2\alpha \theta[/tex]
where;
[tex]\omega _f^2 = \omega_i^2 + 2\alpha \theta\\\\\omega _f^2 = (0) + 2(6)(1.571)\\\\\omega_f^2 = 18.852\\\\\omega_f = 4.34 \ rad/s[/tex]
Thus, the angular velocity of the uniform 50-lb slender rod when rotated 90 degrees is determined as 4.34 rad/s.
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