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標題:
發問:
A wheel on a game show is given an initial angular speed of 1.42 rad/s. It comes to rest after rotating through 0.75 of a turn. Find the average torque exerted on the wheel given that it is a disk of radius 0.56 m and mass 5.4 kg.
最佳解答:
Angular displacement, θ = 0.75 X 2π = 1.5π rad Initial angular speed, ω = 1.42 rad/s Final angular speed, ω’ = 0 By ω’^2 = ω^2 + 2αθ 0 = (1.42)^2 + 2α(1.5π) Average angular acceleration, α = -0.214 rads^-2 As the wheel is a disk, the moment of inertia, I is given by I = 1/2 mr^2 = 1/2 (5.4)(0.56)^2 = 0.84672 kgm^2 By Γ = Iα Torque required, Γ = (0.84672)(-0.214) = -0.181 Nm So, the magnitude of average torque required = 0.181 Nm Alternative method: Moment of inertia of the wheel, I = 0.84672 kgm^2 By conservation of energy, Work done by the torque = rotational K.E. lost Γθ = 1/2 Iω'^2 – 1/2 Iω^2 Γ(1.5π) = 1/2 (0.84672)(1.42)^2 – 0 Torque required, Γ = -0.181 Nm
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College Physics發問:
A wheel on a game show is given an initial angular speed of 1.42 rad/s. It comes to rest after rotating through 0.75 of a turn. Find the average torque exerted on the wheel given that it is a disk of radius 0.56 m and mass 5.4 kg.
最佳解答:
Angular displacement, θ = 0.75 X 2π = 1.5π rad Initial angular speed, ω = 1.42 rad/s Final angular speed, ω’ = 0 By ω’^2 = ω^2 + 2αθ 0 = (1.42)^2 + 2α(1.5π) Average angular acceleration, α = -0.214 rads^-2 As the wheel is a disk, the moment of inertia, I is given by I = 1/2 mr^2 = 1/2 (5.4)(0.56)^2 = 0.84672 kgm^2 By Γ = Iα Torque required, Γ = (0.84672)(-0.214) = -0.181 Nm So, the magnitude of average torque required = 0.181 Nm Alternative method: Moment of inertia of the wheel, I = 0.84672 kgm^2 By conservation of energy, Work done by the torque = rotational K.E. lost Γθ = 1/2 Iω'^2 – 1/2 Iω^2 Γ(1.5π) = 1/2 (0.84672)(1.42)^2 – 0 Torque required, Γ = -0.181 Nm
其他解答:
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