Three particles, each of mass $m$ are placed at the points $(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) a n d (x_{3}, y_{3}, z_{3})$ on the inner surface of a paraboloid of revolution obtained by rotating the parabola, $x^{2} = 4 a y$ about the $y$ -axis. Neglect the mass of the paraboloid. ( $y$ -axis is along the vertical)

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If the particle at $(x_{1}, y_{1}, z_{1})$ slides down the smooth surface, its speed at $O$ is $\sqrt{2 g y_{1}}$

If the paraboloid spins about $O Y$ with an angular speed $ω$ , the kinetic energy of the system will be $2 m a (y_{1} + y_{2} + y_{3}) ω^{2}$ .

If potential energy at $O$ is taken to be zero, the potential energy of the system is $m g (y_{1} + y_{2} + y_{3})$ .

The moment of inertia of the system about the axis of the paraboloid is $I = 4 m a (y_{1} + y_{2} + y_{3})$ .

answer is A, B, C, D.

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Detailed Solution

For any point on the surface of paraboloid, $(x^{2} + z^{2}) = 4 a y$

$I = \sum_{i = 1}^{3} m ({x_{i}}^{2} + {z_{i}}^{2}) = 4 ma (y_{1} + y_{2} + y_{3})$ (distance of $m_{i}$ from $y$ -axis is $\sqrt{{x_{i}}^{2} + {z_{i}}^{2}}$ )
$P . E = mg (y_{1} + y_{2} + y_{3})$
${mgy}_{1} = \frac{1}{2} m {v_{1}}^{2} \Rightarrow v_{1} = \sqrt{2 {gy}_{1}}$
Distance $y$ -mass $m_{p}$ from $y$ -axis, $r_{i} = \sqrt{{x_{i}}^{2} + {z_{i}}^{2}} = \sqrt{4 {ay}_{i}}$