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Thread: [SOLVED] Time period of oscillations?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Time period of oscillations?

    A uniform board of length $\displaystyle L$ and weight $\displaystyle W$ is balanced on a fixed semicircular cylinder of radius $\displaystyle R$ 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 $\displaystyle T = \frac{\pi L}{\sqrt{3gR}}$.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    A uniform board of length $\displaystyle L$ and weight $\displaystyle W$ is balanced on a fixed semicircular cylinder of radius $\displaystyle R$ 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 $\displaystyle T = \frac{\pi L}{\sqrt{3gR}}$.

    When the point of contact is at an angle $\displaystyle \theta$ 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 $\displaystyle I\ddot{\theta} = -Wd$, 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 $\displaystyle \theta$ is small, we can use the approximate value $\displaystyle d\approx R\theta$. 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 $\displaystyle I = \tfrac1{12}ml^2$. So we take $\displaystyle I\approx \frac{WL^2}{12g}$.

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