# [SOLVED] Time period of oscillations?

• May 29th 2009, 09:39 AM
fardeen_gen
[SOLVED] Time period of oscillations?
A uniform board of length $L$ and weight $W$ is balanced on a fixed semicircular cylinder of radius $R$ as shown in the figure. If the plank is tilted slightly from its equilibrium position, then show that time period of oscillations is given by $T = \frac{\pi L}{\sqrt{3gR}}$.
• May 30th 2009, 01:48 AM
Opalg
Quote:

Originally Posted by fardeen_gen
A uniform board of length $L$ and weight $W$ is balanced on a fixed semicircular cylinder of radius $R$ as shown in the figure. If the plank is tilted slightly from its equilibrium position, then show that time period of oscillations is given by $T = \frac{\pi L}{\sqrt{3gR}}$.

When the point of contact is at an angle $\theta$ with the vertical, the forces on the plank are its weight W (at the centre of the plank) and an upwards force N at the point of contact. The angular equation of motion (about the point of contact) is $I\ddot{\theta} = -Wd$, where I is the moment of inertia about the point of contact, and d is the horizontal distance from the centre of the plank to the point of contact.

Assuming that $\theta$ is small, we can use the approximate value $d\approx R\theta$. Also, we can assume that I is the same as the moment of inertia about the centre of the plank. The formula for the moment of inertia of a rod of mass m and length l about its midpoint is $I = \tfrac1{12}ml^2$. So we take $I\approx \frac{WL^2}{12g}$.

Then the (approximate) angular equation of motion is $\frac{WL^2}{12g}\ddot{\theta} = -WR\theta$, or $\ddot{\theta} = -\frac{12Rg}{L^2}\theta$. That is an SHM equation for a motion with period $2\pi/\sqrt{12Rg/L^2} = \pi L/\sqrt{3Rg}$.