Which Equation Determines Angular Displacement for a Rotating Pulley Wheel?

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SUMMARY

The discussion centers on calculating angular displacement for a solid pulley wheel rotating at a constant velocity. The initial angular velocity (Wo) is 6 Rad/s, and the angular acceleration (α) is 0 Rad/s². Two equations are presented: θ = Wo*t + 1/2*α*t² and θ = (Wo + W)/2 * t. The correct angular displacement is determined to be 60 Rads, as the final angular velocity (W) is incorrectly assumed to be 0. The conclusion is that when angular acceleration is zero, the simpler equation θ = Wo*t should be used.

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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
\alpha = 0 (angular acceleration) Rads/s-2
θ = ? (angular displacement) Rads/s
t = 10 sec


Homework Equations



θ=Wo*t+1/2*\alpha*t squared

θ=(Wo+W)/2)*t


The Attempt at a Solution



θ=Wo*t+1/2*\alpha*(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 \alpha in it?

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

Another example is if I decelerate:

Wo = 6 Rads
\alpha =-3Rads
t = 2 sec

θ=Wo*t+1/2*\alpha*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*\alpha*(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 \alpha 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
\alpha =-3Rads
t = 2 sec

θ=Wo*t+1/2*\alpha*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.
 

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