Period of Oscillation of a Meter Stick

[kaydubss]

Homework Statement

Measure the period of oscillation of a meter stick suspended at one end for small amplitudes.
Does T = 2*sqrt(2I/mgL) when I is the rotational inertia about one end? Derive this relation.

Homework Equations

I = rotational inertia
m = mass
g = gravitational acceleration
T = period of oscillation

I=mr^2

T=2*pi*sqrt(L/g)

The Attempt at a Solution

I'm not going to lie, I have no idea where to start. Previous period of oscillation problems with a simple pendulum or a spring/glider system were based off of the formula for period that included 2pi divided by the angular frequency. I don't know where the pi went or what to do with the r value if I was to substitute it in. HELP!

 

[rock.freak667]

I think your equation is missing a π in it.


Anyhow, try considering that at a small angle, what the height is (above the lowest point).

For θ being small, what is cosθ and sinθ equal to?

Now try conserving some energy.

 

[kerimek]

Hello, thank you for your question. To answer your first question, yes, T = 2*sqrt(2I/mgL) when I is the rotational inertia about one end. This relation can be derived using the formula for the period of oscillation of a simple pendulum, T = 2*pi*sqrt(L/g), and the formula for rotational inertia, I = mr^2.

To begin, we can start with the formula for the period of oscillation of a simple pendulum, T = 2*pi*sqrt(L/g). This formula tells us that the period is dependent on the length of the pendulum, L, and the gravitational acceleration, g.

Now, let's consider the rotational inertia of the meter stick. Rotational inertia, I, is a measure of an object's resistance to changes in its rotational motion. For a meter stick suspended at one end, the rotational inertia would be about that end. The formula for rotational inertia is I = mr^2, where m is the mass of the object and r is the distance from the axis of rotation to the object.

Substituting this formula for I into our equation for the period of oscillation, we get T = 2*pi*sqrt(L/g) = 2*pi*sqrt(mr^2/g). Simplifying this further, we get T = 2*pi*sqrt(m/g)*r.

Now, let's focus on the term sqrt(m/g). This term is equivalent to sqrt(1/g)*sqrt(m). Since g is the gravitational acceleration, we can rewrite sqrt(1/g) as 1/sqrt(g). Therefore, the term sqrt(m/g) can be rewritten as sqrt(m)/sqrt(g).

Substituting this back into our equation, we get T = 2*pi*(sqrt(m)/sqrt(g))*r. Simplifying this further, we get T = 2*sqrt(mr^2/g)*r.

Finally, we can replace mr^2 with I, since that is the rotational inertia of our meter stick about one end. This gives us our final equation, T = 2*sqrt(I/g)*r. Since r is the length of our meter stick, we can rewrite this as T = 2*sqrt(I/g)*L.

Therefore, we have derived the relation T = 2*sqrt(2I/mgL) when I is the rotational inertia about one end.

I hope