Vertical acceleration of a rod rotating around a pivot

[yankans]

Homework Statement

A uniform rod of mass 1.6 kg is 6 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance of 6 m from the center of mass of the rod. Initially the rod makes an angle of 58 degrees with the horizontal. The rod is released from rest at an angle of 58 degrees with the horizontal.

What is the magnitude of the vertical acceleration of the center of mass of the rod at the instant the rod is in a horizontal position? The acceleration of gravity is 9.8 m/s^2 and the moment of inertia of the rod about its center of mass is 1/12 mL^2. Answer in units of m/s^2.

Homework Equations

torque = I (moment of inertia) * angular acceleration
tang. acceleration = R*angular acceleration

The Attempt at a Solution

torque = I * ang. acceleration
r mg = 1/12 m l^2 * ang. acc.
6mg = 1/12 m (6^2) * ang. acc.
6mg = 3m* ang. acc.
ang. acc = 2g
tang. acc. = r(ang. acc) = 2(6g)
tang. acc(vertical) = 12g

And it is wrong.

 

[Doc Al]

At the position indicated, what does the vertical component of the acceleration depend on?

Note that you'll need the rotational inertia about the pivot point.

(I assume that the "thin extension" is perpendicular to the rod.)

 

[yankans]

The thin extension is attached to the rod in the same direction as the rod (i.e.) if I drew a picture it would sort of look like a corn dog:
------((((((((((((((
dist. from end of the thin extension to center of mass of rod = 6m
length of rod itself = 6m
so length of the thin extension = 3m

 

[Doc Al]

The thin extension is attached to the rod in the same direction as the rod (i.e.) if I drew a picture it would sort of look like a corn dog:
------((((((((((((((
dist. from end of the thin extension to center of mass of rod = 6m
length of rod itself = 6m
so length of the thin extension = 3m

OK. My previous comments remain (except my guess as to how the extension was attached).

Was the initial angle of the rod 58 degrees above or below the horizontal?

 

[yankans]

above the horizontal

 

[yankans]

it's falling down

 

[Doc Al]

OK, answer my question from post #2.

 

[yankans]

hmm...I guess it would depend on gravity (I originally looked at this problem and thought that the answer was just g, but my physics teacher told me I was wrong), because the total vertical acceleration apparently also (I think) depends on the angular acceleration.

 

[Doc Al]

The Attempt at a Solution

torque = I * ang. acceleration
r mg = 1/12 m l^2 * ang. acc.
6mg = 1/12 m (6^2) * ang. acc.

You are using the moment of inertia about the center of mass; you should be using the moment of inertia about the pivot. (Consider the parallel axis theorem.)

 

[yankans]

Oh wow, I see, problem solved!

Thank you!