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A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?

1. π² m s⁻² and direction along the radius towards the centre
2. π² m s⁻² and direction along the radius away from the centre
3. π² m s⁻² and direction along the tangent to the circle
4. π²/4 m s⁻² and direction along the radius towards the centre

Correct answer: π² m s⁻² and direction along the radius towards the centre

Solution

The stone is undergoing uniform circular motion, so its acceleration is centripetal and directed towards the center of the circle. The centripetal acceleration is given by a = ω²r, where ω = 2π/T and T = time period per revolution = 44/22 = 2 s. Substituting ω = π rad/s and r = 1 m, we get a = π² m/s².

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