Moment of Inertia of Flywheel

[kk727]

Homework Statement

The flywheel of an engine has a moment of inertia 2.50 (kg)m^2 about its rotation axis.
What constant torque is required to bring it up to an angular speed of 400 rev/min in a time of 8.00s, starting from rest?Time (t) =8.00 sec
Ang Velocity (W) = 400 rev/min
Moment of Inertia (I) = 2.50 kg m^2

Homework Equations

W=W_o+αt
T = αI

The Attempt at a Solution

So, I converted the 400 rev/min to rad/sec.
400 * 2pi/60sec = 41.8876 rad/sec

I then plugged everything into the first equation to solve for angular acceleration.
41.8876 rad/sec = 0 + α(8 sec)
α = 5.2358 rad/sec^2

I then found torque using T=αI.
T=(5.2358)(2.50)
T=13.0895 N*m

Now, I thought that was the answer, so I tried looking up the question online to verify it. I found this:
http://answers.yahoo.com/question/index?qid=20101018171042AAmoOu0

I followed the same step process, but my only question is this -

From Yahoo Answers website...w (final) = w (initial) + (c) x time (t)
15pi= 0 + c(8.10x60seconds)
c= 0.0969

Why did that person multiply the time by 60 seconds? I did everything the same as them, except for this. Can someone explain why?

 

[vela]

For no apparent reason. Your solution is correct; his isn't.

 

[kk727]

Hahaha, okay, it totally confused me and made me doubt everything that I did...thanks for the verification! :)