Rotational Motion and Equilibrium

[hbailey]

Homework Statement

A 2.5 kg pulley of radius 0.15m is pivoted about an axis through its center. What constant torque is required for the pulley to reach an angular speed 25rad/s after rotating 3.0 revolutions, starting from rest?

Homework Equations

torque = (mr^2)(angular acceleration)

The Attempt at a Solution

First I solved for time using t= omega/angular speed = 6(pie) rad / 25 rad/s = 0.75s.
Then, I solved for angular acceleration = 33 rad/s

Solving for torque, using the above equation, I got 1.9 m-N

The textbook I have says this is the wrong answer. What have I done?

The book says the answer is 0.47 m-N

 

[e(ho0n3]

First I solved for time using t= omega/angular speed = 6(pie) rad / 25 rad/s = 0.75s.

That seems wrong. Why is omega $6\pi$ radians? Isn't omega the angular speed? What you're calculating is the amount of time it would take for the pulley to rotate $6\pi$ radians at 25 rad/s.

 

[Dick]

In using t=omega/angular speed you are assuming that the angular speed is constant. As it starts from rest, this can't be right.

 

[chaoseverlasting]

Torque/I = angular acceleration. w=0+at, where a=angular accn. 6pie=0.5at^2, using equations of rotational motion. Solve to get torque.

 

[hbailey]

Homework Statement

A 2.5 kg pulley of radius 0.15m is pivoted about an axis through its center. What constant torque is required for the pulley to reach an angular speed 25rad/s after rotating 3.0 revolutions, starting from rest?

Homework Equations

torque = (mr^2)(angular acceleration)

The Attempt at a Solution

First I solved for time using t= omega/angular speed = 6(pie) rad / 25 rad/s = 0.75s.
Then, I solved for angular acceleration = 33 rad/s

Solving for torque, using the above equation, I got 1.9 m-N

The textbook I have says this is the wrong answer. What have I done?

The book says the answer is 0.47 m-N

Oh, instead of omega above, I meant theta.