Question

Three particles of same mass m are fixed to a uniform circular hoop of mass m and radius R at the corner of an equilateral triangle .The hoop is free to rotate in a vertical plane about a point on the circumference opposite to one of the masses .Find the equivalent length of the simple pendulum .

Answer 1

Moment of inertia of all parts

I= mR² + mR² + mR² + mR²

 = 4mR²

Total mass (M)= 4m

Time period, T= 2π√(l/g)

   Here T= 2π √{(Icm+Mα²)/Mgd}

               = 2π √{(4mR²+4mR²)/4mgd}

               = 2π √(2R/g)

Hence equivalent length l= 2R