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A disk is fixed at its centre O and rotates with constant angular velocity omega. A rod AB has end A attached to the disk at a point located R/2 vertically above O. The other end B is connected to a ring that can slide freely along a fixed vertical smooth rod. At the instant when rod AB makes an angle of 30 deg with the vertical, the ring moves upward with speed v. The length of rod AB is L. Find omega.

1. v*sqrt(3)/(2*L)
2. v*sqrt(3)/(2*R)
3. 2*sqrt(3)*v/R
4. 2*v/(L*sqrt(3))

Correct answer: 2*sqrt(3)*v/R

Solution

A is at distance R/2 from O (rotating axis), so speed of A = omega*R/2, directed horizontally. B moves vertically with speed v. Rigid rod constraint: (v_A - v_B). r_hat = 0, where r_hat is unit vector along AB. Rod makes 30 deg with vertical, so r_hat = (sin 30, cos 30) = (1/2, sqrt3/2). v_A = (omega*R/2, 0); v_B = (0, v). (v_A - v_B). r_hat = (omega*R/2)*(1/2) + (0-v)*(sqrt3/2) = 0. => omega*R/4 = v*sqrt3/2 => omega = 2*sqrt3*v/R.

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