Question

A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal,
frictionless pin through one end of the rod. (The moment of inertia of the rod about this
axis is ML2/3.) The rod is released when it makes an angle of 37° with the horizontal.
What is the angular acceleration of the rod at the instant it is released?

Answer 1

Use Newton's Second Law of Rotation
\[\alpha *I=\tau net\]

\[\tau=RF \]
\[\tau= (.6*2*9.8*\cos(37))\]
\[\tau=9.39=I \alpha =\frac{ 1 }{ 3 }ml ^{2}\alpha\]\[\alpha=9.78 rad/\sec ^{2}\]