Which Equation Determines Angular Displacement for a Rotating Pulley Wheel?

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lubo
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Homework Statement


A solid pulley wheel is rotated at a constant velocity for 10 sec's. Find the angular displacement?

Wo (inital angular velocity) = 6Rad/s
W (angular Velocity) = 0 Rads/s
[itex]\alpha[/itex] = 0 (angular acceleration) Rads/s-2
θ = ? (angular displacement) Rads/s
t = 10 sec


Homework Equations



θ=Wo*t+1/2*[itex]\alpha[/itex]*t squared

θ=(Wo+W)/2)*t


The Attempt at a Solution



θ=Wo*t+1/2*[itex]\alpha[/itex]*(t squared)

displacement θ = 6 * 10 + 1/2 * 0 * t = 60 Rads

or θ=(6+0)/2)*10 = 30 Rads.

I cannot tell which one is right as they both should find θ. All I know is that there is an angular acceleration of 0 and 60 is the right answer. Does this mean I have to use an equation with [itex]\alpha[/itex] in it?

I thought I could use anyone I liked as it found θ.

Another example is if I decelerate:

Wo = 6 Rads
[itex]\alpha[/itex] =-3Rads
t = 2 sec

θ=Wo*t+1/2*[itex]\alpha[/itex]*t squared

displacement θ = 6 * 2 + 1/2 * -3 * (2^2) = 6 Rads

or θ=(6+0)/2)*2 = 6 Rads.

Why would this work out or is it just luck? I know the answer to be 6 Rads.

Thank you for any help in advance. :smile:
 
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lubo said:
Wo (inital angular velocity) = 6Rad/s
OK.
W (angular Velocity) = 0 Rads/s
:confused: I thought the speed was constant? W = 6 rad/s.

θ=Wo*t+1/2*[itex]\alpha[/itex]*(t squared)

displacement θ = 6 * 10 + 1/2 * 0 * t = 60 Rads
That's fine.

or θ=(6+0)/2)*10 = 30 Rads.
You are using W = 0, which is incorrect.

I cannot tell which one is right as they both should find θ. All I know is that there is an angular acceleration of 0 and 60 is the right answer. Does this mean I have to use an equation with [itex]\alpha[/itex] in it?

I thought I could use anyone I liked as it found θ.
You can use either one. (Of course, all you really need is θ = ωt, since the speed is constant.) But you have to use the right values.

Another example is if I decelerate:

Wo = 6 Rads
[itex]\alpha[/itex] =-3Rads
t = 2 sec

θ=Wo*t+1/2*[itex]\alpha[/itex]*t squared

displacement θ = 6 * 2 + 1/2 * -3 * (2^2) = 6 Rads

or θ=(6+0)/2)*2 = 6 Rads.

Why would this work out or is it just luck? I know the answer to be 6 Rads.
A little bit of luck. You assumed that the final velocity ω equaled 0 at t = 2 sec. Happens to be true.