A cord is wound around a cylindrical spool that is free to spin around a frictio

Question

A cord is wound around a cylindrical spool that is free to spin around a frictionless axle. The cord is then pulled from the spool with a constant force of 4.5 N. if the spool has a moment of inertia equal to .08 kg m^2 and a radius of .2 m what is the angular acceleration of the spool? A cord is wound around a cylindrical spool that is free to spin around a frictionless axle. The cord is then pulled from the spool with a constant force of 4.5 N. if the spool has a moment of inertia equal to .08 kg m^2 and a radius of .2 m what is the angular acceleration of the spool?

Explanation / Answer

The moment of inertia for the given system, I = mr2

Substituting for I and radius r in the above formula gives the mass, m = I / r2

                                                                             m = 2 kg

Also given that F = 4.5 N = ma

Substituting for mass m, we can evaluate the linear acceleration

                           a = 2.25 m/s2

Given the linear acceleration, angular acceleration, = a/r

                          = 11.25 rad/s2

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