Finding Angular Acceleration of Uniform Rod

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Homework Statement



A uniform rod of length 0.750 m and mass 1.20 kg is pivoted at one end by a smooth pin as shown below. The rod is released from the vertical position and given a slight nudge to release it from the vertical position of unstable equilibrium.

[PLAIN]http://img175.imageshack.us/img175/9820/angacc40pc.png

When the rod is horizontal:

(a) Calculate its angular acceleration.

(b) Calculate the x and the y components of the acceleration of the centre of mass.

The Attempt at a Solution



First I focus on part (a). I have already obtained the angular velocity:

The potential energy of the system relative to the reference configuration is MgL/2 because the center of mass of the rod is at a height L/2 away from its position in the reference configuration. Conservation of mechanical energy for the system is:

Ki+Ui = Kf + Uf

0+½MgL=½Iω²+0

[tex]\omega = \sqrt{\frac{MgL}{\frac{1}{3}ML^2}} = \sqrt{\frac{3g}{L}}[/tex]

using the given values I get ω=6.26 rad/s, which is the correct velocity.

Now I want to use the equation [tex]\alpha = \frac{\Delta \omega}{t}[/tex] to find the angular acceleration. Here's where I'm stuck. Could anyone please show me how to find the time for this equation?

Correct answer to part (a) is 19.62 rad/s² and for part (b) it is -14.7 m/s²(x-component) and -7.36 m/s² (y-component)
 
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No, you can not use

[tex] \alpha = \frac{\Delta \omega}{t}[/tex],
as it is valid for uniform acceleration only, and you do not know if the acceleration is uniform; neither you know the time.

There is a formula betwee torque, angular acceleration, and moment of inertia. Use that.

ehild
 
ehild said:
No, you can not use

[tex] \alpha = \frac{\Delta \omega}{t}[/tex],
as it is valid for uniform acceleration only, and you do not know if the acceleration is uniform; neither you know the time.

There is a formula betwee torque, angular acceleration, and moment of inertia. Use that.

ehild

Thank you very much, I got it. Now could you explain how to deal with part (b)? How do we calculate the x (or y) component of the acceleration of the centre of mass? I'm not quite sure what that even means since I don't have any worked examples in my textbook.