A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum . How does it differ fro the time period calculated using the formula for a simple pendulum?

A hollow sphere of radius 2 cm is attached to an 18 cm long thread to

1.	A hollow sphere of radius 2 cm is attached to an 18 cm long thread to make a pendulum. Find the time period of oscillation of this pendulum . How does it differ fro the time period calculated using the formula for a simple pendulum?
Answer» Solution :According to Energy equation `momega(0.2)(1-costheta)+1/2Iomega^2=C`……….1 Againg I=2/3m(0.02)^2+m(0.2)^2` `=2/3m(0.0004)+m(0.4)` `=m[0.0008+0.12]` `=m(0.1208)/3)` where lrarr moent of inertia about the POINT of suspension A. From equation differenting and putting the value of l in equation 1 ils `d/(dt)[MG(0.2)(1-costheta)+1/20.1203/3momega^2]` `d/(dt) (c)` `rarr mg(0.2)sintheta(dtheta)/(dt)+1/2(0.1208/3)m^2omega(domega)/(dt)=0` `rarr 2sintheta=0.1208/3 alpha` `[because g=10m/s^2]` `alpha/theta=6/0.1208` `=omega^2=58` `rarr omega=7.63` So, `T=(2pi)/3` `=0.89sec` For simple PENDULUM `T=2pi sqrt(0.19/10) `=0.86sec` `% more =(0.89-0.86)/0.89=0.3% `:.` It is about 0.3% larger than the calculated value.