Asked by t
A rope is wrapped around the rim of a large uniform solid disk of mass 265 kg and radius 2.90 m. The horizontal disk is made to rotate by pulling on the rope with a constant force of 195 N. If the disk starts from rest, what is its angular speed in rev/s at the end of 2.85 s?
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Torque τ = Fr = (195 N)(2.90 m) = 565.5 N·m.
Moment of inertia for a solid disk I = (1/2)MR^2 = 0.5(265 kg)(2.90 m)^2 = 0.5·265·8.41 = 1115.25 kg·m^2.
Angular acceleration α = τ/I = 565.5/1115.25 = 0.5075 rad/s^2.
After t = 2.85 s from rest, ω = αt = 0.5075·2.85 = 1.446 rad/s.
Convert to revolutions per second: f = ω/(2π) = 1.446/(2π) = 0.230 rev/s.
Answer: 0.230 rev/s (about 1.45 rad/s).
Moment of inertia for a solid disk I = (1/2)MR^2 = 0.5(265 kg)(2.90 m)^2 = 0.5·265·8.41 = 1115.25 kg·m^2.
Angular acceleration α = τ/I = 565.5/1115.25 = 0.5075 rad/s^2.
After t = 2.85 s from rest, ω = αt = 0.5075·2.85 = 1.446 rad/s.
Convert to revolutions per second: f = ω/(2π) = 1.446/(2π) = 0.230 rev/s.
Answer: 0.230 rev/s (about 1.45 rad/s).
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