Three particles, each of mass `m` are placed an the points `(x_(1),y_(1),z_(1)),(x_(2),y_(2),z_(2))` and `(x_(3),y_(3),z_(3))` on the inner surface of a paraboloid of revolution obtained by rotating the parabola `x^(2)=4ay` about the `y`-axis. Neglected the mass of the paraboloid. (`y`-axis. is along the vertical (a) the moment of inertia of the system about the axis of the paraboloid is `I=4ma(y_(1)+y_(2)+y_(3))`

Options

A. the moment of inertia of the system about the axis of the paraboloid is `I=4ma(y_1+y_(2)+y_(3))`
B. if potential energy at `O` is taken to be zero, the potential energy of the system is `mg(y_(1)+y_(2)+y_(3))`
C. if the particle at`(x_(1),y_(1),z_(1))` slides down the smooth surface, its speed at `O` is `sqrt(2gy_(1))`
D. if the parabola spins about `OY` with an angular speed, `omega`, the kinetic energy of the system will be `2ma(y_(1)+y_(2)+y_(3))omega^(2)`.

Answer

A, B, C, D

Explanation

For any point on the sufrace of paraboloid,`(x^(2)+z^(2))=4ay`
`I=sum_(i=1)^(3)m(x_(i)^(2)+z_(i)^(2))` (distance of `m_(1)` from `y`-axis is)
`sqrt(x_(i)^(2)+z_(i)^(2))=4ma(y_(1)+y_(2)+y_(3))`
`mg(x_(1)+x_(2)+x_(3))=mg(y_(1)+y_(2)+y_(3))`
`mgy_(1)=1/2mupsilon_(1)^(2)rArrupsilon_(1)sqrt(2gy_(1))`
distance `y`-mass `m_(p)` from `y`-axis
`r_(i)=sqrt(x_(i)^(2)+z_(i)^(2))=sqrt(4ay_(i))`
`KE=1/2 m omega^(2)(r_(1)^(2)+r_(2)^(2)+r_(3)^(2))`
`=1/2momega^(2)4a(y_(1)+y_(2)+y_(3))`