A circular table is rotating with an angular velocity of ω rad / s about its axis (see figure). 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 w.r.t. the table at the time ball leaves the table is x √2 ω m / s, where the value of x is _________.

Solution

Physics sir · 3 steps

In the table's frame only the centrifugal force pushes along the groove, so integrate it from 1 m out to 3 m.

1. Rotating frame. Work in the frame of the table. Along the smooth groove the only force is the centrifugal force mω² r, directed outward. The Coriolis force acts sideways and is balanced by the groove wall, so it does no work along the groove.
2. Set up acceleration. The radial acceleration is ω² r, which grows as the ball moves out. Write it as vdv/dr = ω² r and integrate from r = 1 m, where v = 0, to r = 3 m.

   ∫₀ᵛ v dv = ω²∫₁³ r dr ⇒ v² = ω²(3² - 1²) = 8ω²

3. So. v = √8 ω = 2√2 ω m s⁻¹. Comparing with x√2 ω gives x = 2.

The idea to keep. A bead in a smooth radial groove speeds up as v² = ω²(r² - r₀²). The centrifugal force depends on r, so integrate with v dv/dr; never treat it as a constant force.

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