A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is fre

Question

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 210 N applied to its edge causes the wheel to have an angular acceleration of 1.124 rad/s2. What is the moment of inertia of the wheel? kg. m2 What is the mass of the wheel? kg If the wheel starts from rest, what is its angular velocity after 4.00 s have elapsed, assuming the force is acting during that time? rad/s Need Help? A 225-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s? (State the magnitude of the force.) N Need Help? A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 400 mile trip in a typical midsize car produces about 1.10 times 109 J of energy. How fast would a 14 kg flywheel with a radius of 0.35 m have to rotate in order to store this much energy? Give your answer in rev/min. rev/min

Explanation / Answer

7 (a) torque = I*alpha F*r = I*alpha I = 210*0.33/(1.124) I = 61.65 (b) I = mr^2/2 m = 2I/r^2 = 2*61.65/(0.33)^2 m = 132.32 (c) w = alpha*t = 1.124*4 = 4.496 8 F*r = (mr^2/2)*(0.700*2*pi/2) F = (225*1.5/2)*(0.700*2*pi/2) F = 371.1 N 9 E = Iw^2/2 w = sqrt(2E/I) w = sqrt(1.1*10^9/(14*0.35^2/2)) w = 35816.181173 rad/s w = 35816.181173*60/2*pi rev/min w = 1368077.35 rev/min

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