A large grinding wheel in the shape of a solid cylinder of radius0.330 m is free
ID: 1745796 • 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 appliedto its edge causes the wheel to have an angular acceleration of0.913 rad/s2. (a) What is the moment of inertia of thewheel?1kg·m2
(b) What is the mass of the wheel?
2 kg
(c) If the wheel starts from rest, what is its angular velocityafter 4.40 s have elapsed, assuming theforce is acting during that time?
3 rad/s (a) What is the moment of inertia of thewheel?
1kg·m2
(b) What is the mass of the wheel?
2 kg
(c) If the wheel starts from rest, what is its angular velocityafter 4.40 s have elapsed, assuming theforce is acting during that time?
3 rad/s
Explanation / Answer
Given that the radius of the solid cylinder is R = 0.330 m Thetangential force is F = 290 N The angular accelaration is = 0.913rad/s2 ----------------------------------------------------------------------- The torque on thecylinder is = I R*F= I where I is the moment of inertia of the wheel then I = R*F / = -----------Kg.m2 The moment of inertia of thesolid cyllinderof mass M is I = (1/2)MR2 M= 2*I / R2 = ------------ kg From the equation ofmotion the final angular veloity with the initial angularveloctiy 1 = 0 is 2 = 1 + *t =0 + *t = *t = ---------- rad/s M= 2*I / R2 = ------------ kg From the equation ofmotion the final angular veloity with the initial angularveloctiy 1 = 0 is 2 = 1 + *t =0 + *t = *t = ---------- rad/sRelated Questions
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