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A toy top with a spool of diameter 4.8 cm has a moment of inertia of 3.5×10−5 kg⋅m2 about its rotation axis. To get the top spinning, its string is pulled with a tension of 0.33 N. How long does it take for the top to complete the first five revolutions? The string is long enough that it is wrapped around the top more than five turns

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Answer:

t = 0.7452secs

Explanation:

Given that, radius = 4.8cm/2 = 2.4cm = 0.024m

Force (F) = 0.33N

Moment of inertia (I) = 3.5 × 10^-5kg/m²

1. Torque applied = Fr = 0.33×0.024

= 0.00792

= 7.92 × 10^-3N/m

2. Angular acceleration = Torque/inertia = 7.92×10^-2/3.5×10^-5

= 226.29rad/s²

3. Five rotations: @ = 5×2π

= 10π, let t be the time taken, then, angular acceleration = 1/2@t²

Therefore, t = √2@/angular acceleration

t = √2×10π/226.29

t = 0.7452secs.

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