Understanding Torque in Wheel Stopping: A Quick Guide

[ProSk8ter474]

There's a question that asks: if a wheel with 1.56 m/rad radius which reached a final velocity of 200 rad/min(or 10/3 rad/s) were slowed down with a constant torque of 10,000 NM being applied, how long would it take to stop the wheel. I can't seem to find how torque is related to this problem in anyway. I can solve it by doing the following:

Angular Displacement = (W^2-Wo^2)/(2*Angular Acceleration)
= (11.11rad^2/s^2)/(4.44rad/s^2)
= 2.5 rad
Angular Displacement = (t/2)(Wo+W)
=(2.5 rad) = (t/2)*(10/3rad/s)
=(2.5 rad)/(3.33rad/s)
= .75s = (t/2)
t = 1.5 s

But it may not be correct because I didn't use the radius or torque in that problem. Please respond with your ideas on how they could be applied to this problem.

 

[Chi Meson]

for now: T= torque A= angular acceleration I= moment of inertia

T=IA so A=T/I

now use the angular acceleration in an angular kentmatics problem with final angular velocity =0

 

[Doc Al]

... I can solve it by doing the following:

Angular Displacement = (W^2-Wo^2)/(2*Angular Acceleration)
= (11.11rad^2/s^2)/(4.44rad/s^2)
= 2.5 rad

You are for some reason assuming a value (of 2.22 rad/s^2) for the angular acceleration. Where did that come from?

If you had the angular acceleration, you wouldn't bother calculating the angular displacement--you'd immediately calculate the time given that you know the Δω.

But it may not be correct because I didn't use the radius or torque in that problem.

That should give you a hint that something's wrong!

As Chi Meson explained, you need to apply Newton's 2nd law for rotation to find α, α = Τ/I.