Work done by the torque on a pulley

[IAmPat]

Homework Statement

The pulley in the illustration is a uniform disk of mass 2.40 kg and radius 0.220 m. The block applies a constant torque to the pulley, which is free to rotate without friction, resulting in an angular acceleration of magnitude 0.180 rad/s² for the pulley. As the block falls 0.500 m, how much work does it do on the pulley?

Homework Equations

Net Torque = Inertia * $$\alpha$$
Work = Torque * $$\Delta$$X

The Attempt at a Solution

Mass of Pulley = 2.40kg
Radius = 0.220m
$$\alpha$$ = 0.180 rad/s²

I = $$\sum$$m * r²
I = 2.40kg * (0.220)²
I = 0.1161

$$\sum$$T = I * $$\alpha$$ = (0.11616) *(0.180)
T = 0.02091


$$\Delta$$ theta = (Change in X)/R = 0.500/0.220= 2.27


W = T * (Change in Theta) = (0.02091)(2.27)
W = 0.0475 J


I must have made a mistake somewhere.

 

[tiny-tim]

Hi IAmPat!

The moment of inertia of a uniform disc is not mr².

Learn the commonly-occurring moments of inertia at http://en.wikipedia.org/wiki/List_of_moment_of_inertia_tensors"

 

[IAmPat]

Hi IAmPat!

The moment of inertia of a uniform disc is not mr².

Learn the commonly-occurring moments of inertia at http://en.wikipedia.org/wiki/List_of_moment_of_inertia_tensors"

Found that the moment of inertia of a uniform disc is 1/2*mr²

Plugged that in and got the correct answer. Thanks.