A uniform rod is 2.0 m long. The rod is pivoted around a horizontal, frictionmuch less pin with one end. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the immediate it is released?

I simply supplied

$$sin(30) = frac9.8a_centripetal$$

Then I related

$$a_rad = a_centripetal$$

Is this right? I am looking at University rwandachamber.org:

$$a_rad = v^2 / r$$

Where 9.8 is gravity. But I obtained none of the answers in the multiple option ... so I have to be doing wrong. Also I haven"t offered the radius. Any suggestions would certainly be advantageous...

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edited Oct 28 "12 at 12:23

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Almeans begin through a nice clear diagram/sketch of the problem. It all follows from there. Here is a Free Body Diagram I created you.


Then you have actually (the long thorough way):

Sum of the forces on body amounts to mass times acceleration at the facility of gravity. $sum_i vecF_i = m veca_C $

$$ A_x = m a_x \ A_y - m g = m a_y $$

Sum of torques about center of gravity equates to minute of inertia times angular acceleration.$sum_i left(vecM_i + (vecr_i-vecr_C) imesvecF_i ight) = I_C vecalpha $

$$ A_x fracL2 sin( heta) - A_y fracL2 cos( heta) = I_C ddot heta $$

Acceleration of suggest A must be zero. $veca_A = veca_C + vecalpha imes(vecr_A-vecr_C) + vecomega imes(vecv_A-vecv_C) $

$$ a_x + fracL2 sin( heta) ddot heta + fracL2 dot heta^2 cos( heta) =0 \ a_y - fracL2 cos( heta) ddot heta + fracL2 dot heta^2 sin( heta) =0 $$

Now you deserve to settle for $a_x$, $a_y$ from 3. and also usage those in 1. to acquire $A_x$,$A_y$. Finally usage 2. to deal with for $ddot heta$

Or execute the shortcut of finding the used torque on A and also using it to the efficient minute of inertia about the pivot $I_A = I_C + m left(fracL2 ight)^2 $ to get

$$ ddot heta = fracm g fracL2 cos heta I_C + m left(fracL2 ight)^2 $$