A mass of 0.450 kg rotates at constant speed with a period of 1.45 s at a radius R of 0.140 m in the apparatus used in this laboratory. What is the rotation period for a mass of 0.550 kg at the same radius

1.603 s

Explanation:

Given that

Initial mass, = 0.45 kg

Initial period, = 1.45 s

Initial radius, = 0.14 m

Final mass, = 0.55 kg

Final period, = ?

Final radios, = 0.14 m

Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that

m₁r₁ω₁² = m₂r₂ω2²

Where, ω = 2π/T, on substituting, we have

0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²

0.45 / 1.45² = 0.550 / T₂²

T₂² = 0.550 * 1.45² / 0.45

T₂² = 2.56972

T₂ = √2.56972

T₂ = 1.603 sec

a.) is the correct answer.

explanation:

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