Question

Solved already! Please check my work?

Answer 1

A thin rod (length = 3.00 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward.
- What is the magnitude of the angular acceleration of the rod just before it strikes the floor?


My answer:
w = \[\frac{ \sqrt{2*9.81*3}}{ 3}\]
2.56 rad/s

2.56^2 = 3.14A
A= 2.08

I have gotten 2.08 everytime I have solved, but it says the correct answer is 3.27.
Anyonesee what I did wrong?

Answer 2

here we have to apply the second equation of rigid bodies. Such equation, is equivalent to the subsequent scalar equation:
\[I\dot \omega = - mgL\sin \theta \]
where \(I\) is the moment of inertia of the object at the top of the rod with respect to the rotation axis (hinge), \(m\) is the mass of such object, and \(\theta\) is like below:

Answer 3

since:
\[I = m\frac{{{L^2}}}{2}\]
after a substitution, we get:
\[\dot \omega = - \frac{g}{L}\sin \theta \Rightarrow \left| {\dot \omega } \right| = \frac{g}{L}\sin \theta \]
Now, I replace \(\theta= \pi/2\), so I get:
\[\left| {\dot \omega } \right| = \frac{g}{L} = \frac{{9.81}}{3} = ...?\]
please complete the computation above