Question

A circular are `(AB)` of thin wire frame of radius `R` and mass `M` makes an angle of `90^(@)` at the origin.The centre of mass of the arc lies at

A. `[0,(2//pi)R]`
B. `[0,(sqrt(2)//pi)R]`
C. `[0,(2sqrt(2)//pi)R]`
D. `[0,(4//pi)R]`

Answer 1

By symmetry
`x_(cm)=0`
`dm=lambdaR d theta`
`y_(cm)=(intydm)/(intdm)=(1)/(M)intRcosthetalambdaRd theta=(R^(2)lambda)/(M)int_(-pi//4)^(+pi//4)costheta d theta`
`=(R^(2)lambda)/(Rxxlambdaxx(pi//2))[2sin"(pi)/(4)]=(2R)/(pi)2(1)/(sqrt(2))=(2sqrt(2)R)/(pi)`