# Pivoting Rod

1. Mar 11, 2004

### jstealth03

here's the question:

A uniform rod of length 1.15 m is attached to a frictionless pivot at one end. It is released from rest from an angle of 27 degrees above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass.

So far I've only been able to figure out that I have to use Newton's Second Law of rotation, F x r = I x alpha. How do i derive that to figure out acceleration? Do i break down F into ma, and cancel the m's on each side that result from inertia?

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2. Mar 11, 2004

### Staff: Mentor

Since the rod is constrained to rotate about the hinge, all you need is Newton's 2nd law for rotation. That's: &Tau; = I &alpha;.

The only force creating a torque about the pivot is the object's weight. You'll also need to know the rotational inertia of the rod.

So, solve for &alpha;. The linear acceleration is r&alpha;.

Do things algebraically and you'll find that things will cancel nicely.