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Math Help - [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 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}}.
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  2. #2
    MHF Contributor
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    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}.
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