A uniform rod of mass M and length L is pivoted at one end. The rod is left to freely rotate under the influence of its own weight. Find its angular acceleration α when it makes an angle 30° with the vertical axis. Solve for M=1 Kg, L=1 m, take g=10 m s-2. Your answer in X.X rad s-2. Hint: Find the center of mass for the rod, and calculate the torque, then apply Newton as τ= Ι·α
As we know that the weight Mg of the rod will act at the centre
of mass that is L/2 distance from
the pivot end.
Now torque about the pivot end = Mg*(L/2)Sin30 = (1*9.81)*(1/2)Sin30 = 2.4525 Nm
We know that
torque(T) = I
where I is the moment of inertia = ML2/3 = (1)*12 /3 = 0.333 kg-m2
and is the angular acceleration
T = I
2.4525 = 0.333*
= 7.3575 rad/s2
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