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 from the time period calculated using the formula for a simple pendulum?

Answer

A, B, C, D

Explanation

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.