Respuesta :
Answer:
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
The rotation period for a mass of 0.550 kg at the same radius is 1.603 s
Calculation of the radius:
Since
Initial mass = 0.45 kg
Initial period = 1.45 s
Initial radius = 0.14 m
Final mass = 0.55 kg
Final radios = 0.14 m
Now the following formulas should be used.
m₁r₁ω₁² = m₂r₂ω2²
here ω = 2π/T
So,
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
hence, The rotation period for a mass of 0.550 kg at the same radius is 1.603 s
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