On a smooth horizontal fixed plane ABCD, a mass m1 = 0.1 kg moves in a circular path of radius r = 1 m. It is connected by an ideal string through a smooth hole in the plane to a mass m2 = 1/sqrt(2) kg hanging below, which itself moves in a horizontal circle of radius r = 1 m with speed v2

Question

On a smooth horizontal fixed plane ABCD, a mass m1 = 0.1 kg moves in a circular path of radius r = 1 m. It is connected by an ideal string through a smooth hole in the plane to a mass m2 = 1/sqrt(2) kg hanging below, which itself moves in a horizontal circle of radius r = 1 m with speed v2 = sqrt(10) m/s. Given g = 10 m/s², find the speed v1 of m1.

Accepted Answer

(A) sqrt(5) m/s. For m2 (conical pendulum below the plane): string length from hole to m2 traces a cone. The radius of m2's circle = r = 1 m, and the string length is L. sin(theta) = r/L, cos(theta) = h/L. Equations: T*sin(theta) = m2*v2²/r and T*cos(theta) = m2*g. Squaring and adding: T² = m2²*(v2⁴/r² + g²) = (1/2)*((10)²/1 + 100) = (1/2)*200 = 100. T = 10 N. For m1: T = m1*v1²/r => 10 = 0.1*v1²/1 => v1² = 100 => v1 = 10 m/s. Alternatively, checking answer sqrt(5): if T = 0.1*5/1 = 0.5 N which does not balance. Let me recompute: T² = (1/sqrt(2))² * ((sqrt(10))⁴/1² + 10²) = (1/2)*(100 + 100) = 100. T = 10 N. v1 = sqrt(T*r/m1) = sqrt(10*1/0.1) = sqrt(100) = 10 m/s.

Solution

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