Pivoting Stick Angular Acceleration and Force Calculation

[anap40]

[SOLVED]Pivoting Stick

Homework Statement

A uniform stick of mass M = 1.3 kg and length L = 2.3 m is pivoted at one end. It is held horizontally and released. Assume the pivot is frictionless. Find the angular acceleration (in rad/s) of the stick immediately after it is released.

Continuation: Find the magnitude in Newtons of the force Fo exerted on the stick by the pivot immediately after it is released.

http://img134.imageshack.us/img134/1377/prob06azg9.gif

Homework Equations

torque=Ia

The Attempt at a Solution

I set T=Ia
t=1.15(1.3)(9.81)=14.67
I=mr^2=1.3(1.15)^2=1.72

14.67=1.72a
a=8.53rad/s(s)

It tells me that is the wrong answer

For the second part, shouldn't the force be 0?(0is not the right answer)

 

[Doc Al]

I set T=Ia
t=1.15(1.3)(9.81)=14.67

OK.

I=mr^2=1.3(1.15)^2=1.72

Not OK. What's the rotational inertia of a thin rod about one end? (When analyzing rotational motion, you can't treat an extended body as if its mass were concentrated at its center of mass.)

 

[anap40]

Ah, thanks, using I =1/3mL^2 I get the correct answer.

So for the second part is the force=9.81(1.3)-6.39(1.15)(1.3)

or in other words the force on the rod it it was not connecting to a pivot point, minus the acceleration of the rod while it is connected to the pivot point?

EDIT: i just put in my answer as caclulated above and it was correct.

thanks Doc Al for the Help

 

[Doc Al]

Cool. The way to think of the second part is just to apply Newton's 2nd law to the vertical direction:
F - mg = ma
(where a is the acceleration of the center of mass, which you can figure out from the first answer)