A large grinding wheel in the shape of a solid cylinder of radius0.330 m is free
ID: 1679546 • Letter: A
Question
A large grinding wheel in the shape of a solid cylinder of radius0.330 m is free to rotate on a frictionless, vertical axle. Aconstant tangential force of 290 N applied to its edge causes the wheel tohave an angular acceleration of 0.896 rad/s2. (a) What is the moment of inertia of the wheel?kg·m2
(b) What is the mass of the wheel?
kg
(c) If the wheel starts from rest, what is its angular velocityafter 5.10 s have elapsed,assuming the force is acting during that time?
rad/s (a) What is the moment of inertia of the wheel?
kg·m2
(b) What is the mass of the wheel?
kg
(c) If the wheel starts from rest, what is its angular velocityafter 5.10 s have elapsed,assuming the force is acting during that time?
rad/s
Explanation / Answer
Radius R = 0.33 m Tangential force F = 290 N angular acceleration = 0.896 rad/s2 Torque = FR = 96.66 N m (a). moment of inertia of the wheel I = / =107.88 rad / s 2 (b). We know I = ( 1/2) M R 2 Mass of the wheel M= ( 2 ) I / R 2 = 1981.26 kg (c). Intitial angular speed = 0 rad / s Angular speed after t = 5.1s is ' = ? we know from the relation ' = + t = 4.5696 rad / s = 1981.26 kg (c). Intitial angular speed = 0 rad / s Angular speed after t = 5.1s is ' = ? we know from the relation ' = + t = 4.5696 rad / sRelated Questions
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