 #1
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Homework Statement:
 A rod of length 50cm is pivoted at one end. It is raised such that it makes an angle 30 degrees from the horizontal as shown and released from rest. It's angular speed when it passes through the horizontal will be
Relevant Equations:

wf^2  wi^2 = 2*alpha*theta,
w=angular velocity
Torque =F*r
Torque =I*alpha
The centre of mass of the rod would be at the middle of the rod i.e. at
l/2=[50*10^(2)]/2
The force responsible for torque will be acting downwards = mg
The Torque = mg*l/2*sin(30) =mg*l/4
We know that Torque=I*alpha
Hence alpha = mg*l/(4*I)
Moment of inertia of rod about the end= ml^2/12 + ml^2/4 (parallel axis theorem) =ml^2/3
Hence alpha=3mgl/(4ml^2) =3g/4l
Now wf^2 wi^2 =2*alpha *theta
=2*3g/4l*pi/6 since theta =pi/6 since it starts from rest wi=0
Hence wf^2 = 2*3*10/(4*0.5)*pi/6
Hence wf= 3.96
But the answer is sqrt(30). Apparently they haven't considered the angle pi/6. Where am I wrong?
l/2=[50*10^(2)]/2
The force responsible for torque will be acting downwards = mg
The Torque = mg*l/2*sin(30) =mg*l/4
We know that Torque=I*alpha
Hence alpha = mg*l/(4*I)
Moment of inertia of rod about the end= ml^2/12 + ml^2/4 (parallel axis theorem) =ml^2/3
Hence alpha=3mgl/(4ml^2) =3g/4l
Now wf^2 wi^2 =2*alpha *theta
=2*3g/4l*pi/6 since theta =pi/6 since it starts from rest wi=0
Hence wf^2 = 2*3*10/(4*0.5)*pi/6
Hence wf= 3.96
But the answer is sqrt(30). Apparently they haven't considered the angle pi/6. Where am I wrong?