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Hinged Rod- angular speed

  • Thread starter Crystal037
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  • #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?
 

Answers and Replies

  • #2
haruspex
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The Torque = mg*l/2*sin(30) =mg*l/4
Two problems with that.
The initial position is 30 degrees above horizontal, not 30 degrees from vertical.
The angle will change as the rod falls, so the angular acceleration increases.

Can you think of an approach that avoids torques and accelerations?
 

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