Rotational MotionHard
Question
Three particles, each of mass m are placed at the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) on the inner surface of a paraboloid of revolution obtained by rotating the parabola, x2 = 4ay about the y-axis. Neglect the mass of the paraboloid. (y-axis is along the vertical)
Options
A.The moment of inertia of the system about the axis of the paraboloid is I = 4ma (y1 + y2 +y3).
B.If potential energy at O is taken to be zero, the potential energy of the system ismg (y1 + y2 + y3)
C.If the particle at (x1, y1, z1) slides down the smooth surface, its speed at O is 
D.If the paraboloid spins about OY with an angular speed ω the kinetic energy of the system will be ma (y1 + y2 + y3)ω2
Solution
For any point on the surface of paraboloid, (x2 + z2) = 4ay
(A) I =
m(x12 + z12) (distance of mi from y-axis is 
) = 4ma (y1 + y2 + y3).
(B) mg (x1 + x2 + x3) = mg(y1 + y2 + y3)
(C) mg x1 =
mv12 ⇒ v1 = 
(D) Distance y-mass mp from y-axis, ri =
KE =
mω2(r12 + r22 + r32) = 
mω2 4a(y1 + y2 + y3)
(A) I =
m(x12 + z12) (distance of mi from y-axis is ) = 4ma (y1 + y2 + y3).(B) mg (x1 + x2 + x3) = mg(y1 + y2 + y3)
(C) mg x1 =
mv12 ⇒ v1 = (D) Distance y-mass mp from y-axis, ri =
KE =
mω2(r12 + r22 + r32) = mω2 4a(y1 + y2 + y3)Create a free account to view solution
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