Which Equation Determines Angular Displacement for a Rotating Pulley Wheel?

AI Thread Summary
To determine the angular displacement of a rotating pulley wheel, the key equation is θ = Wo * t + 1/2 * α * t², where Wo is the initial angular velocity, α is the angular acceleration, and t is time. In this case, with a constant velocity of 6 Rad/s over 10 seconds and zero angular acceleration, the correct calculation yields an angular displacement of 60 Rads. The alternative equation θ = (Wo + W)/2 * t can also be used, but it requires the correct final angular velocity, which is not zero in this scenario. The discussion highlights the importance of using the right values in equations, especially when angular acceleration is zero. Ultimately, the correct approach confirms that θ = 60 Rads is accurate for constant velocity conditions.
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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