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.