Most Important Selected Qs for JEE AdvancedPhysicsRotational Motion
Three particles, each of mass m are placed at 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=4 a y about the y -axis. Neglect the mass of the paraboloid. ( y -axis is along the vertical)
Options
- AThe moment of inertia of the system about the axis of the paraboloid is I=4 m a (y₁+y₂+y₃ ) .
- BIf potential energy at O is taken to be zero, the potential energy of the system is m g (y₁+y₂+y₃ ) .
- CIf the particle at ( x ₁, y ₁, z ₁ ) slides down the smooth surface, its speed at O is 2 gy ₁
- DIf the paraboloid spins about OY with an angular speed , the kinetic energy of the system will be 2 ma ( y ₁+
Correct answer
D. If the paraboloid spins about OY with an angular speed , the kinetic energy of the system will be 2 ma ( y ₁+
Step-by-step solution
For any point on the surface of paraboloid, (x^2+z^2 )=4 a y (A) I = _ i =1 ^3 ~m ( x ₁^2+ z ₁^2 ) (distance of m _ i from y -axis is . x _ i ^2+ z _ i ^2 )=4 ma ( y ₁+ y ₂+ y ₃ ) . (B) m g (x₁+x₂+x₃ )=m g (y₁+y₂+y₃ ) (C) m g x₁= 1 2 mv ₁^2 v ₁= 2 gy ₁ (D) Distance y -mass m_p from y-axis, r_i= x_i^2+z_i^2 = 4 a_i KE = 1 2 ~m ^2 ( r ₁^2+ r ₂^2+ r ₃^2 )= 1 2 ~m ^2 4 a ( y ₁+ y ₂+ y ₃ )