A solid cylinder of mass 3.0 kg and radius 1.0 m is rotating about its axis with a speed of 40 rad s−1. Calculate the torque which must be applied to bring it to rest in 10s. What would be the power required?
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MOI of cylinder = I₂ = M R²/2 = 3.0 * 1.0² /2 = 1.5 kg-m²
ω = 40 rad/s ω = 0 rad/s t = 10 s
Angular acceleration = α = (40 - 0)/t = 4 rad/sec²
Torque = I₂ α = 1.5 * 4 = 6 N-m
ω = 40 rad/s ω = 0 rad/s t = 10 s
Angular acceleration = α = (40 - 0)/t = 4 rad/sec²
Torque = I₂ α = 1.5 * 4 = 6 N-m
kvnmurty:
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What is the power
Instantaneous rotational Energy = E = 1/2 I ω^2... Instantaneous power = dE/dt = I ω dω/dt = I ω α . Here α is constant. We need to find average power from t=0 to t=10s, or for ω = 40 rad/s to ω=0.
ω = ω0 - alpha t = 40 - 4 t. The average ω is ω0/2 = 20 rad/s. So average power is = I *ω0/2* α = 1.5 * 20 * 4 = 120 Watts.
sorry i missed that.
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