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## 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?