# What tension must a 47.9 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 3.37 m/s?

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• if vertical

T = m*g+m*V^2/r = 1*(9.806+3.37^2/0.479) = 33.5 N

if horizontal

T = m*V^2/r = 3.37^2/0.479) = 23.7 N

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• To support the weight of the stone, the string must be at angle below horizontal. Let θ be the angle. The vertical component is equal to the weight of the stone. The horizontal component is equal to the centripetal force. The length of the string is 0.479 meter.

Weight = 1 * 9.8 = 9.8 N

Fc = 1 * 3.37^2 ÷ 0.479 = 11.3569 ÷ 0.479

This is approximately 23.7 N. Since these two forces are perpendicular to each other, we can use the following equation to determine the magnitude of the tension.

T = √(Vertcal^2 + Horixontal^2)

T = √[(9.8^2) + (11.3569 ÷ 0.479)^2]

This is approximately 25.7 N. To determine the angle, use the following equation.

Tan θ = Weight ÷ Centripetal force

Tan θ = 9.8 ÷ (11.3569 ÷ 0.479) = 4.6942 ÷ 11.3569

The angle is approximately 22.5˚ below horizontal.

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• assuming a horizontal circle, so gravity is not relevant...

f = mV²/r = 1•3.37²/0.479 = 23.7 N

Centripetal force f = mV²/r = mrω²

ω is angular velocity in radians/sec

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