A thin homogeneous rod with a length of 1.2 m and a mass of 300 g, located vertically, can rotate around

A thin homogeneous rod with a length of 1.2 m and a mass of 300 g, located vertically, can rotate around a horizontal axis passing through the upper end, perpendicular to the rod. The rod is deflected from the vertical axis at an angle of 60 ° and released. With what angular velocity will the rod pass the equilibrium position?

Let’s write the formulas for solving the problem:
1.h = l – l * cos a;
2.E = E = m * g * h = m * g * (l – l * cos a);
3. E = m * v ^ 2/2;
Let’s use the law of conservation of energy:
m * g * (l – l * cos a) = m * v ^ 2/2;
u = √ (2 * g * (l – l * cos a));
w = v / R = √ (2 * g * (l – l * cos a)) / l = √ (2 * g * (1 – l * cos a)) / l;
w = 2.9 rad / s.

Answer: w = 2.9 rad / s.