A bead of mass m stays at point P(a, b) on a wire bent in the shape of a parabola y = 4Cx² and rotating with angular speed ω (see figure). The value of ω is (neglect friction) :

Solution

Physics sir · 4 steps

The normal force tilts with the wire, so match its tilt to the slope of the parabola at P.

1. Forces on the bead. With no friction, only gravity mg (down) and the normal force N (perpendicular to the wire) act. The bead moves in a horizontal circle of radius a, so the net force must point horizontally towards the axis.
2. Resolve N. If the tangent at P makes angle θ with the horizontal, the normal makes the same angle θ with the vertical. Vertically, N cos θ = mg. Horizontally, N sin θ = mω² a. Dividing gives tan θ = (ω² a)/g.
3. Slope at P. From y = 4Cx², dy/dx = 8Cx. At x = a, tan θ = 8Ca.

   (ω² a)/g = 8Ca ⇒ ω² = 8gC

4. So. ω = √8gC = 2√2gC, which is option (a). Notice that a cancels, so this ω holds the bead at any point of the wire.

The idea to keep. For a bead on a smooth rotating curve, tan θ = (ω² r)/g, where tan θ is the slope dy/dx of the curve at that point. Set them equal and solve.

Answer from the NTA final answer key. Solution by PrepWiser.