Angular acceleration on a rod about an axis

[Leigh10590]

Homework Statement

A long, 1 kg uniform rod and 1.4 m length, is supported at the left end by a horizontal axis into the page and perpendicular to the rod. The right end is connected to the ceiling by a thin vertical thread so that the rod is horizontal.
The thread is burned by a match.
Find the force exerted on the rod by the axis immediatley after the thread breaks.
When the rod is at an angle of 80 degrees with the horizontal, find the angular velocity of the rod.

Homework Equations

the moment of Inertia of the rod is (ML^2)/3

The Attempt at a Solution

I solved for the force of the horizontal axis and the string to both be 4.9 N upward when at equilibrium. The angular acceleration of the rod is 10.5 rad/s^2. I also solved for the translational acceleration of the center of mass to be 7.35 m/s^2. I have no clue how to do the last two parts of the problem. Someone please help me!

 

[LowlyPion]

Homework Statement

A long, 1 kg uniform rod and 1.4 m length, is supported at the left end by a horizontal axis into the page and perpendicular to the rod. The right end is connected to the ceiling by a thin vertical thread so that the rod is horizontal.
The thread is burned by a match.
Find the force exerted on the rod by the axis immediatley after the thread breaks.
When the rod is at an angle of 80 degrees with the horizontal, find the angular velocity of the rod.

Homework Equations

the moment of Inertia of the rod is (ML^2)/3

The Attempt at a Solution

I solved for the force of the horizontal axis and the string to both be 4.9 N upward when at equilibrium. The angular acceleration of the rod is 10.5 rad/s^2. I also solved for the translational acceleration of the center of mass to be 7.35 m/s^2. I have no clue how to do the last two parts of the problem. Someone please help me!

You might want to remember the conservation of energy.

$$\Delta PE = \Delta KE = \Delta KE_{rotational}$$

$$m*g*\Delta h = \frac{I\omega^2}{2}$$

Your drop in height then of the center of mass would translate into the rotational kinetic energy. Your height of interest is when the height has dropped to the angle of 80 degrees perhaps?