A reel of mass 2.00 kg and radius 25.0 cm is free to rotate about its center. At t=0 a tangential force is applied to the reel as shown
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The moment of inertia of a solid cylinder rotating about its central axis is given by the formula I = (1/2) * m * r^2, where m is the mass of the reel and r is the radius. Substitute m = 2.00 kg and r = 0.25 m into the formula: I = (1/2) * 2.00 kg * (0.25 m)^2 I Show more…
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A solid disk of mass 500 grams and a radius of 10.0 cm is free to rotate about its center. At t=0, a tangential force is applied to the reel. The angular speed changes with time according to the equation w(t) = 3.00t^1.5, where w is in rad/s. a) At t=4.00s, find the torque being exerted on the disk, in N*m. b) Find the magnitude of the tangential force on the disk, in newtons. c) Find the number of revolutions undergone by the reel in 4.00 seconds. d) Find the amount of work done by the force in rotating the disk during the 4.00s, in joules.
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A force of $P=20 \mathrm{N}$ is applied to the cable, which causes the 175 -kg reel to turn since it is resting on the two rollers $A$ and $B$ of the dispenser. Determine the angular velocity of the reel after it has made two revolutions starting from rest. Neglect the mass of the rollers and the mass of the cable. The radius of gyration of the reel about its center axis is $k_{G}=0.42 \mathrm{m}$
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