[SOLVED] Time period of oscillations?

fardeen_gen · #1 $Old$ May 29th, 2009, 09:39 AM

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 $T = \frac{\pi L}{\sqrt{3gR}}$ .

Opalg · #2 $Old$ May 30th, 2009, 01:48 AM

Quote:

Originally Posted by fardeen_gen $View Post$

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 $2\pi/\sqrt{12Rg/L^2} = \pi L/\sqrt{3Rg}$ .