Calculating Torque and Linear Acceleration of a Rotating Rod

[ScienceGeek24]

-1/2Lmg=1/3 M L^2(alpha) ? is this right?

 

[gneill]

-1/2Lmg=1/3 M L^2(w)^2 ?

There's no w^2. Newton's second law for rotation is $\tau = I \alpha$

 

[ScienceGeek24]

now, how do i cancel the Ls? one has L and the other one has L^2?? even if i manage to cancel them out i don't get the right answer.

 

[gneill]

now, how do i cancel the Ls? one has L and the other one has L^2?? even if i manage to cancel them out i don't get the right answer.

You're not done yet. Solve for $\alpha$. What expression do you get?

 

[ScienceGeek24]

α=-1/2g-1/3L

 

[gneill]

α=-1/2g-1/3L

Something went wrong there... You started with

-1/2Lmg=1/3 M L^2(alpha) (assuming m is really M)

How did you end up with two terms? Try that again.

 

[ScienceGeek24]

sorry! i mmeant α=-1/2g/1/3 I got the answer! 14.7 m/s^s thanks man!

 

[gneill]

sorry! i mmeant α=-1/2g/1/3 I got the answer! 14.7 m/s^s thanks man!

$\alpha = -\frac{3}{2}\frac{g}{L}$

You've still got the L in the denominator to deal with. It goes away when you calculate the linear acceleration of the endpoint... $a = \alpha L$.

 

[ScienceGeek24]

oh got it! thanks bro!