A particle of mass m is suspended from point O and undergoes circular motion in horizontal plane as conical pendulum as shown in figure.

Options

A. Angular momentum of particle about point of suspension does not remains constant.
B. Angular momentum of particle about centre of circle remains constant.
C. Average force during half rotation is `(2mg tan thetha)/pi`
D. Average torque about axis OC during half rotation is zero

Answer

A, B, C, D

Explanation

`tau_(av)xxDeltat=L_(2)-L_(1)=DeltaL`
`vec(L)=-Lcosthetahat(i)+Lsinthetahat(j), vec(L)_(2)=Lcosthetahat(i)+Lsinthetahat(j)`
`vec(L)_(2)-vec(L)_(1)=(2Lcosthetahat(i))`
magnitude is same but direction of angular momentum is continuously changing So `vec(L)` is not constant (B) about centre is constant (c)`Tcos theta=mg......(1)`
`Tsintheta=momega^(2)lsintheta..........(2)`
`momega^(2)lcostheta=mg`
`omega^(2)=g/(lcostheta), omega=sqrt(g/(lcostheta))`
`V=omega(lsintheta)=sqrt((gl)/(costheta)) sintheta`
`F_(av)xxDeltat=DeltaP=P_(2)-P_(1)`
`mV=msinthetasqrt((gl)/(costheta))`
`F_(av)=(P_(2)-P_(1))/(Deltat)=((2msintheta)sqrt((gl)/(costheta)))/(pi//omega)`
`F_(av)=(2msintheta)/pi sqrt((gl)/(costheta))xxsqrt(g/(lcostheta))Rightarrow F_(av)=(2mg tan theta)/pi`
L will be constant about line OC as shown in diagram