# A simple harmonic problem

1. May 19, 2015

### MMONISM

1. The problem statement, all variables and given/known data
A L = 1.48 m long metal rod is pivoted from a point one third of the way along its length. The rod has a mass of 0.208 kg. What period will the rod oscillate with when released from a small angle?

2. Relevant equations
$T = 2 \pi \sqrt{\frac{L}{g}}$

3. The attempt at a solution
should I think the rod as a point mass about it's centre mass? and use centre mass to calculate the period by the formula?
This is my approach:

$\frac{1}{2}L-\frac{1}{3}L$ is the length of the string
$T = 2 \pi \sqrt{\frac{L}{g}} = 2 \pi \sqrt{\frac{\frac{L}{6}}{g}} = 2 \pi \sqrt{0.025} = 0.997 s$

2. May 19, 2015

### Staff: Mentor

That approach is too simple. Consider a rod pivoted very close to the center of mass - your formula would give a very short period, but the actual period would be very long.

Think of torque and angular momentum.

3. May 20, 2015

### MMONISM

distance from pivot to centre mass is L/6
torque(mg) = sin(theta)*(L/6)*mg
also torque = I*angular acceleration.
I am confused here, as w in simple harmonic motion is angular frequency and in the torque formula it is angular velocity? are they the same?

4. May 21, 2015

### Staff: Mentor

They are not, and I don't see where an $\omega$ would appear in the formula for torque. In general: if you would get the same symbol for different meanings, use a different symbol or indices (point 4 here).