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Frictional torque!! will rate super duper LIFESAVER!

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Question:

First of all I can't tell the difference between applied torque and frictional torque...

So, here's the question:

The combination of an applied force and a frictional force produces a constant total torque of 35.6 N·m on a wheel rotating about a fixed axis. The applied force acts for 6.07 s, during this time the angular speed of the wheel increases from 0 to 10.2 rad/s. The applied force is then removed, and the wheel comes to rest in 60.2 s.

(a) Find the moment of inertia of the wheel.
correct check mark

kg·m²
(b) Find the magnitude of the frictional torque.
wrong check mark

N·m
(c) Find the total number of revolutions of the wheel.
rev

Answers:

LAPHROAIG

Oracle

Karma Points: 13,361

(Santa Fe Community College-New Mexico)

Date Posted: 5/8/2008 1:18:04 PM Status: Live

Asker's Rating: Lifesaver

Response:

I'm only helping you out because you said I would get super duper lifesaver. Lol

First we know that I= 21.19kgm² and we also know τ=Iα=|(21.19kgm²)(-10.2rad/s/60.2s)|= 3.59 Nm.
c) Δθ= 1/2αt²=1/2(10.2rad/s/6.07s)(6.07s)²= 30.957 rads and then Δθ=ωt+1/2αt²=(10.2)(60.2s)+1/2(-10.2rad/s/60.2s)(60.2s)²=307.02 rads, so the total is about 338 rads or 338rad(1rev/2πrad)= 53.8 revs.

"IGNORE THE POST BELOW MINE. HE COPIED MY POST TO TRY TO GET FREE KARMA POINTS."

Koontz's Comment:

thanks!! they don't have that as an option, but you are definitely a super duper lifesaver in my book!

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