Rotational simple harmonic motion occurs when there is a linear restoring torque

Question

Rotational simple harmonic motion occurs when there is a linear restoring torque given by T(r) = -K times theta,where theta is the angular displacement from equilibrium. The period of the resulting rotational oscillation is given in analogy with equation T = 2pi times the square root of (m/k) by T = 2pi times the square root of(I/k). Derive equation T = 2pi times the square root of (mL/kg) = 2pi times the square root of (L/g) by treating the simple pendulum as a rotating system with the suspension point as the pivot. Note that for small angles, sin of theta = theta. (Please show me every step of how u eventually get the new equation)

Explanation / Answer

when the pendulum moves to one side, the restoring force on it is given by -mgsin(-ve sign as it acts opposite to the direction of motion).So the restoring torque about the pivot is given by- -mgsinL(where L is the length of the pendulum)

So, -mgsinL=I

-mgsinL=mL2

So, =-gsin/l=-g/l

So, =g/L (as =-2 )

So, T=2/=2L/g

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