What is the angular acceleration of a uniform rod when a string is released?

[sebasalekhine7]

A uniform rod has a length of 2m. It is hinged to a wall at the left end and held horizontally by a vertical massless string at the right end. What is the angular acceleration of the rod the moment the string is released?

 

[rpc]

angular accel = tangential accel (in this case gravity) / radius

 

[Curious3141]

Easiest way : [tex]\tau = I\alpha[/tex], which means torque = moment of inertia*angular acceleration.

What is the moment of inertia of a rod of uniform length about the end ?

What is the torque exerted by the weight of the rod about the end ?

 

[sebasalekhine7]

I think that [tex]I=1/3(ML^2)[/tex]

 

[Curious3141]

I think that [tex]I=\frac{1}{3}(ML^2)[/tex]

Correct. You don't have to know the derivation, but it's fairly simple using a little calculus.

What's the torque about the end ?

 

[sebasalekhine7]

Ok I got it, thanks, [tex]\tau = I\alpha[/tex] when [tex]\tau =1/2mgL[/tex] and therefore [tex]\frac{1}{3}(ML^2)\alpha=1/2mgL[/tex] and angular acceleration ends up being 7.35.

You see, the problems I post here are the ones I am studying for the next physics science league, is a competition between schools here in NJ.

 

[Curious3141]

Excellent, you got it (except both your m's need to be the same case) !

 

[drnikitin]

A uniform rod has a length of 2m. It is hinged to a wall at the left end and held horizontally by a vertical massless string at the right end. What is the angular acceleration of the rod the moment the string is released?

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