Question

A flat, rigid object oscillates as a physical pendulum in simple harmonic motion with a frequency f. The mass of the pendulum is m, and the pivot point is a distance d from the center of mass. What is the moment of inertia of the pendulum about its pivot point? (Use any variable or symbol stated above along with the following as necessary: g.)

Answer 1

o = 2t is a pivot point e = position of centre of max U OA=It is a point about . which it rotates o to When it rotates of through shall google (= force due to pivot point We take torque about caris OA. -- Magnitude of torque TI Force Xperpendicular . distance from OA- ...111= Exo + mg xd sine : mg T= rng deine L. It becomes ave because torque P. T = IK Acceleration tries to bring back moment of inertia mass, M to the a = 1/1 = -mgd sine mean position.de. 6=0% gangewas beration tries to x = - nad y fire Now t to execute simple harmonic omotion, must be proportion to ao.. J. e. (2) 2Go if G is very small then Sino no So, ... a = -mgd x 0 - 0 motion: we know that, in Sinple. Harmonic @=.-w²o - 0. equating .reg - and egg We nga : We 201f nn f = mgd 21f2 mga YIT?f

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