Find the magnitude of the initial acceleration of the rod's center of mass

[fruitl00p]

Homework Statement

A uniform rod of length 1.15m is attached to a frictionless pivot at one end. It is released from rest an angle theta=17.0 degrees above the horizontal.

Find the magnitude of the initial acceleration of the rod's center of mass

Homework Equations

torque=moment of inertia*angular acceleration
torque=length*force(perpendicular)

I=1/3 ML^2

The Attempt at a Solution

I set the torque about pivot equal to the moment of inertia about pivot times the angular acceleration to find the angular acceleration. Then I found the linear acceleration by multiplying the found angular acceleration by the length of rod. So my work was...

L/2*(Mg)cos17=1/3*ML^2*(angular acceleration)
3/2*(g/L)*cos17=(angular acceleration)
12.22 rad/s^2= (angular acceleration)

a=(angular acceleration)*L= 14.05 m/s^2

However the answer I got is wrong. (Unfortunately, I do not have access to the correct answer) What step did I miss?

 

[andrevdh]

... L/2*(Mg)cos17=1/3*ML^2*(angular acceleration)

This should be the sin not the cos.

The com will then experience only a tangential acceleration:

$$a_t = r \alpha$$

 

[fruitl00p]

I replaced the cos with the sin17 in my equation and I just tried to solve but I was told my answer was wrong. And I did exactly what you suggested. I wonder why that was wrong? Did I need to do something else?

 

[Mentz114]

Have you used the correct angle ? 17 deg above the horizontal is 90 + 17 from the vertical.

 

[hage567]

If you are trying to find the tangential acceleration of the centre of mass, wouldn't it be $$a_t = \alpha(L/2)$$, since the centre of mass will be at half the length of a uniform rod? Just a thought.

 

[fruitl00p]

Mentz114, thank you for mentioning that! I looked through some notes and saw I need to add 90 degrees. However, I just redid the problem, using the info you told me and what andrevdh told me, and oddly enough, I got the same answer!

hage567, I will try to find the initial acceleration with L/2 instead of L.

 

[fruitl00p]

Thanks everyone, I got the right answer now.