(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform rod with of weight of 10 kg and length of 5.8 m is pivoted at its center and a small weight of mass 5.15 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod. The system is released from rest at a 37° angle. There are no external forces. What is the angular acceleration just after it is released?

2. Relevant equations

Torque = Angular Acceleration times Inertia

Torque = Force times Distance from pivot

Inertia of a uniformly weighted bar = 1/12 M L^2

I = Icm + M D^2

3. The attempt at a solution

Since the bar is pivoting about it's center, the net torque of the bar itself is zero. (One side balances the other out.)

The torque of the weight is 2.9 x 50.52 = 146.51

146.51 = I times angular acceleration.

The inertia of the bar = 1/12 (10) 5.8^2 = 28.033

146.5 / 28.033 = 5.226

This is wrong. Clearly I need to calculate the Inertia for the Weight, not just the bar, but since the problem says to neglect the size of the weight, I assume you have to treat it as part of the bar.

How do you calculate the inertia for a non-uniform bar?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Angular Acceleration of a rod

**Physics Forums | Science Articles, Homework Help, Discussion**