Question

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. Hint: Find the center of mass for the rod, and calculate the torque, then apply Newton as τ= Ι·α

Answer #1

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.

Consider the following diagram .

Now torque about the pivot end = Mg*(L/2)Sin30 = (1*10)*(1/2)Sin30
= 2.5 Nm

We know that

torque(T) = I

where I is the moment of inertia = ML^{2}/3 =
(1)*1^{2} /3 = 0.333 kg-m^{2}

and is the angular acceleration

Now

T = I

2.5 = 0.333*

= 7.5 rad/s^{2}

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