A large grinding wheel in the shape of asolid cylinder of radius 0.330 m is free

Question

A large grinding wheel in the shape of asolid cylinder of radius 0.330 m is free to rotate on africtionless, vertical axle. A constant tangential force of 290 Napplied to its edge causes the wheel to have an angularacceleration of 1.162 rad/s2. (a) What is the moment ofinertia of the wheel? answer in kg·m2

(b) What is the mass of the wheel? answer in kg

(c) If the wheel starts from rest, what is its angular velocityafter 6.00 s have elapsed, assuming the force is acting during thattime? answer in rad/s (a) What is the moment ofinertia of the wheel? answer in kg·m2

(b) What is the mass of the wheel? answer in kg

(c) If the wheel starts from rest, what is its angular velocityafter 6.00 s have elapsed, assuming the force is acting during thattime? answer in rad/s

Explanation / Answer

Part A) torque=force*radius=moment_of_inertia*angular_acceleration =F*r=I* 290*.330=I*1.162 I=82.4 kg*m2 Part B) the moment of inertia of a cylinder is I=.5*mr282.4=.5*m*.3302 m=1513 kg Part C) =0+t =0+1.162*6 =6.97 rad/s

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