A uniform rod, pivoted at its upper end, hangs vertically. It is...

Let the mass of the uniform rod be $M and its length be$L$. The moment of inertia of the rod about the pivot is$I = \dfrac{ML^2}{3}$.<br><br>When the rod is at an angle$\theta = 37^\circ$with the vertical, the torque about the pivot is provided by the weight of the rod acting at its center of mass:<br><br>$\tau = Mg \dfrac{L}{2} \sin 37^\circ$<br><br>Using$\tau = I\alpha$, the angular acceleration$\alpha$is:<br><br>$\dfrac{ML^2}{3} \alpha = Mg \dfrac{L}{2} \left(\dfrac{3}{5}\right)

$\alpha =$\dfrac{3g}{2L} \times \dfrac{3}{5} = \dfrac{9g}{10L} = \dfrac{0.9g}{L}

The tangential acceleration of the particle at the tip of the rod is:

$a_t = \alpha L = 0.9g$

To find the angular velocity $\omega$ at $\theta = 37^\circ$, we use the conservation of mechanical energy. The rod is released from rest at $\theta_0 = 60^\circ$. The decrease in potential energy is equal to the gain in kinetic energy:

$\Delta U = Mg$\dfrac{L}{2} (\cos$37^\circ - \cos 60^\circ) =$\dfrac{1}{2}$I \omega^2$

$Mg$\dfrac{L}{2} \left( \dfrac{4}{5} - \dfrac{1}{2} \right) = \dfrac{1}{2} \left( \dfrac{ML^2}{3} \right) \omega^2$<br><br>$Mg $\dfrac{L}{2} \left( \dfrac{3}{10} \right) = \dfrac{ML^2}{6} \omega^2$

\dfrac{3g}{20} = \dfrac{L}{6} \omega^2 \Rightarrow \omega^2 = $\dfrac{18g}{20L} = \dfrac{0.9g}{L}<br><br>The radial (centripetal) acceleration of the particle at the tip is:<br><br>$a_r = \omega^2 L = 0.9g$<br><br>The net acceleration of the particle of mass$dm$is the vector sum of its tangential and radial accelerations:<br><br>$a = $\sqrt{a_t^2 + a_r^2} = \sqrt{(0.9g)^2 + (0.9g)^2} =$0.$9\sqrt{2}$g$<br><br>According to Newton's second law, the magnitude of the net force acting on the particle is:<br><br>$F = dm \times a = 0.$9\sqrt{2}$\,dm\,g$<br><br>Answer:$0.$9\sqrt{2}$\,dm\,g