Could someone help me with this question and show the work a

Could someone help me with this question and show the work as well?

A uniform rod is 36 in long and has a mass of 20 lbm. It is pinned at point A, a frictionless pivot. The rod is released from a rest position. 45^degree from the horizontal. The top of an undeflected ideal spring is located at point B to contact the tip of the rod when the rod is horizontal. The spring has a constant of 525 lbf/in. What is the velocity of the rod\'s tip when the ri just becomes horizontal? 12 ft/sec 14 ft/sec 19 ft/sec 29 ft/sec

Solution

Here, we will use the conservation of energy to determine the speed of the tip of the rod. As the rod falls down the work done by gravity is changed to kinetic energy.

So we can write: mgH = 0.5 I²

here, H is the height by which the centre of mass comes down, I is the inertia of the rod and is the angular velocity of the rod when it becomes horizontal.

That is, 20(32,2) x 1.5 Cos45 = 0.5 x 20 x 3² .x ²/ 3

or, 30² = 683.065

or = 4.772

Therefore the velocity of the tip of the rod when the rod becomes horizontal = 14.3 ft/sec

Hence option B is the required answer.