The radial acceleration is provided by the net radial force T – mg cos θ , while the tangential acceleration is provided by mg sin θ . It is more convenient to work with torque about the support since the radial force gives zero torque. Torque τ about the support is entirely provided by the tangental component of force τ = – L ( mg sin θ ) ( . ) This is the restoring torque that tends to reduce angular displacement — hence the negative sign.
By Newton’s law of rotational motion, τ = I α ( . ) where I is the moment of inertia of the system about the support and α is the angular acceleration. Thus, I α = – m g sin θ L ( . ) Or, α = sin m g L I θ − ( .
) We can simplify Eq. ( . ) if we assume that the displacement θ is small. We know that sin θ can be expressed as, sin ± ...
! ! θ θ θ θ = − + ( . ) where θ is in radians.
Now if θ is small, sin θ can be approximated by θ and Eq. ( . ) can then be written as, α θ = − mgL I ( . ) In Table .
, we have listed the angle θ in degrees, its equivalent in radians, and the value of the function sin θ . From this table it can be seen that for θ as large as degrees, sin θ is nearly the same as θ expressed in radians . Table . sin θθθθθ as ma function of angle θθθθθ (degrees) (radians) sin Equation ( .
) is mathematically, identical to Eq. ( . ) except that the variable is angular displacement. Hence we have proved that for small q, the motion of the bob is simple harmonic.
From Eqs. ( . ) and ( . ), ω = mgL I and I mgL = ( .
) Now since the string of the simple pendulum is massless, the moment of inertia I is simply mL . Eq. ( . ) then gives the well-known formula for time period of a simple