Question

A particle of mass m is initially at rest in a smooth groove in a table as shown. Initially value of $\theta$ is 37°.

Now table starts rotating with angular velocity $\omega = \frac{6}{\sqrt{7}}$ rad/s. Find speed of particle (in m/s) when angle $\theta$ becomes 53° (R = 4 m)

Answer 1

~4.8 m/s

Answer 2

In the rotating frame the effective energy (Jacobi constant) is ½ m v² – ½ m ω²(R sinθ)² = constant.

With the particle initially at rest (θ = 37°) the constant is –½ m ω²(R sin37)².

At θ = 53°, equate energies to get ½ m v² = ½ m ω²R²(sin²53 – sin²37), hence

v = ωR√(sin²53 – sin²37).

Substitute ω = 6/√7, R = 4, sin37 ≈ 0.6018, sin53 ≈ 0.7986 to get v ≈ 4.8 m/s.