A circular table is rotating with an angular velocity of ω rad/s about its axis. There is a smooth groove along a radial direction on the table. A steel ball is gently placed at a distance of 1 m on the groove. All the surfaces are smooth. If the radius of the table is 3 m, the radial velocity of the ball with respect to the table at the time ball leaves the table is x√2ω m/s, where the value of x i $s ______.$ Image

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G1 Physics · Accepted Answer

Image In the rotating frame, the outward radial acceleration is: a(r) = ω²r Using: a = v(dv/dr) we get: v dv = ω²r dr Integrating from: r = 1 m, v = 0 to: r = 3 m, v = v ∫₀ᵛ v dv = ω²∫₁³ r dr v²/2 = ω²(9 - 1)/2 v² = 8ω² Therefore: v = 2√2 ω Given: v = x√2ω Hence: x = 2