angular velocityharmonic-oscillatorrotational-dynamicstorque
For the oscillation of a torsion pendulum (a mechanical motion), the time period is given by
$T=2\pi\sqrt{\frac{I}{C}}$ which is a result of the angular acceleration $\alpha=\frac{d^2\theta}{dt^2}=-(\frac{C}{I})\theta$ where $C$ is the restoring couple of the string.
Do we relate $T=\frac{2\pi}{\omega}$ and $\alpha$ for finding the time period of torsion pendulum? I can't really understand how they arrived at $2\pi\sqrt{\frac{I}{C}}$ ?
Gravity exerts a constant force downwards. If you apply a second constant force sideways you are in effect rotating the gravitational force:
where the angle $\theta$ is given by:
$$ \tan\theta = \frac{F_\text{ext}}{F_g} $$
The modulus of the net force is given by the usual Pythagoras rule:
$$ F_\text{net}^2 = F_g^2 + F_\text{ext}^2 $$
So if you imagine mentally rotating the whole experiment an angle $\theta$ anticlockwise you would have the pendulum handing vertically downwards in an effective gravitational field of $F_\text{net}$. This should make it obvious how the period changes.
The difference is that the ball bearing is rolling on a circular track instead of sliding. When analyzing the motion we must take into account the moment of inertia of the ball bearing.
In a simple pendulum the bob rotates about the pivot point. This is equivalent to sliding (without friction) on the circular track. But the ball-bearing pendulum is rotating about its own centre as it also rotates about the imaginary pivot point at the centre of the circular track.
For a certain amplitude of swing, there is a fixed amount of energy. At the lowest point the kinetic energy of the ball bearing is divided between linear motion of the CM and rotation about the CM, whereas for the simple pendulum bob it all goes into linear motion of the CM. So the linear speed of the CM of the ball bearing is slower than that of the simple pendulum bob, resulting in a longer period for the ball bearing. When we come to measure the period, it is only the back-and-forth motion of the CM which we measure; we ignore the rotation of the ball bearing.
According to Oscillation of a rolling sphere in a bowl, the period of small oscillations is
$$T=2\pi \sqrt{\frac{7R}{5g}}=1.1832T_0$$
where $T_0$ is the period of a simple pendulum of the same length $R$. In the lecture $T_0=1.85s$ so we should expect $T=2.19s$. We can only assume the remaining difference with the measured value of $2.27s$ is due to other errors - eg the measurement of $R$.
Best Answer
There are lots of different examples of oscillatory systems that have essentially the same mathematical form. Let's start by just looking at one type of differential equation:
$a = \frac{d^2 x}{dt^2} = -\omega^2 x$
This equation has a general solution (you can check this)
$x(t) = A \sin (\omega t + \phi)$
which oscillates with a period of $T=2\pi/\omega$ since the system will be in exactly the same state at any time $t$ and $t + 2\pi/\omega$. So now if we find physical systems that are described by an acceleration equation that looks like this, we know exactly what the solution is.
For example, in a mass spring system we would have
$ma=-kx \rightarrow \frac{d^2 x}{dt^2} = -\left(\frac{k}{m}\right) x$
which is exactly the same form as we had before with $\sqrt{\frac{k}{m}}=\omega$, so plugging this into our equation for period we get $T=2\pi\sqrt{\frac{m}{k}} $
The same applies for your torsion pendulum, you switch the position $x$ with angular position $\theta$ (this doesn't change how to solve the differential equation) and have
$\frac{d^2\theta}{dt^2} = -\left(\frac{C}{I}\right) \theta = -\omega^2 \theta$
as your differential equation with $\omega = \sqrt{\frac{C}{I}}$, it has the same solution, and plugging this $\omega$ into your period formula you get
$T=2\pi \sqrt{\frac{I}{C}}$
This shows that once you solve the general form of one type of differential equation you can apply it to any other type of system that has the same form for its equation (this is also used in circuits, perturbations to an orbit, or pretty much anything else that is slightly pushed away from a stable equilibrium position)