Solve Uniform Wire Question: Center of Mass & Torque | QuantumNinja

[Tom McCurdy]

Problem here:
http://www.quantumninja.com/hw/random/problem4.jpg

I was trying to figure out how to go about this problem. So far I have come up with

center of mass=
\frac{L}{4}cos\frac{\beta}{2}

Torque=t
\sum t=I\alpha

 

[Davorak]

Are you sure this is right
cm = \frac{L}{4}cos\frac{\beta}{2}
If \beta goes to zero then the cos will go to one and you will be left with \frac{L}{4} which does not seem right to me for a stright rod.

 

[Tom McCurdy]

I thought that was right

 

[Tom McCurdy]

Actually maybe it would be
L/4*cos(b/2)+L/2
center of mass
\frac{L}{4}cos\frac{b}{2}+\frac{L}{2}

 

[Davorak]

I am sorry you are right. If beta goes to zero the rod becomes lengh of L/2 and the center of mass would the be a L/4.

 

[Tom McCurdy]

Which one is right the first or 2nd one