# Pivoting rod

1. Nov 10, 2005

### Punchlinegirl

A uniform rod of length 1.29 m is attached to a frictionless pivot at one end. It is released from rest from an angle q = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass.
$$I= (1/2)ML^2$$
I= (1/2)(1.29^2), since mass isn't given I just used 1 kg.
I=.832 kg*m^2
$$\tau= rFsin \theta$$ =.645(9.8)(sin 21)= 2.27 Nm
$$\tau = I \alpha$$
2.27= .832 $$\alpha$$
$$\alpha$$ = 2.73 rad/s^2
$$a= \alpha * r$$
a= 2.73 * .645
a= 1.76 m/s^2
This isn't right.. can someone help me?

2. Nov 10, 2005

### lightgrav

This is not a uniform DISK, it is a uniform ROD.
Look up in a table the Inertia
(unless you want to calculate it yourself, using integration)
is it 1/3 M L^2 , or is it 1/12 M L^2 ?

Also, the angle theta that you take the sine of
is the angle that sweeps from the r_vector direction
to the F_vector direction ... as in the right-hand-rule?

(gravity is down, not sideways! )

3. Nov 10, 2005

### Punchlinegirl

I got it. Thanks