How Do You Calculate Average Angular Acceleration?

AI Thread Summary
To calculate the average angular acceleration of a rolling wheel, use the formula (final angular speed - initial angular speed) divided by time. The problem states the wheel starts at 7.0 rad/s and ends at 11.0 rad/s after 3 complete revolutions, which corresponds to an angular displacement of 6π radians. The solution involves using rotational kinematics, specifically the equation ω_f^2 = ω_i^2 + 2αθ, to find angular acceleration without needing time. Additionally, when determining the ratio of final to initial angular momentum, the relationship L = Iω is crucial, where I is the moment of inertia, leading to the conclusion that the ratio is simply the ratio of the angular speeds. Understanding these concepts is essential for solving angular motion problems effectively.
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Homework Statement


A rolling wheel moving to the right initially has an angular speed of 7.0 rad/s but after making 3
complete revolutions the wheel has a final angular speed of 11.0 rad/s. What is the magnitude and direction of the average angular acceleration of the wheel?


Homework Equations


(w2-w1)/(t2-t1) or maybe I need to use that equation and some rotational kinematics equations, not positive.



The Attempt at a Solution


I tried saying my time was 3 seconds but I am having trouble identifying the time. Not really sure where to start with this problem.
 
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The wheel makes 3 complete revolutions ... so the angular displacement is:
You are right, you need kinematics. You know initial velocity, final velocity, and displacement: what is the equation that gives you the acceleration (hint: it does not have time in it)?
 
Awesome ty,

11^2 = 7^2 +2a(6pi)
a = 6pi
 
New question, well a continuation question:

For the speeds given in problem #6 above, what is the ratio of the wheel‟s final angular
momentum to the initial angular momentum?

L = rmvsin(theta)

I know the answer is 1.6 which is 11/7 however, I am not sure why that is. The question gave me a change in theta of 6pi and two speeds. Okay, so wouldn't dividing the speeds give you the ratio of the initial speed and final speed, not the momentum? Or is the speed also the momentum? I am confused.
 
vysero said:
a = 6pi
Not quite. Try that last step again.
For the speeds given in problem #6 above, what is the ratio of the wheel‟s final angular
momentum to the initial angular momentum?

L = rmvsin(theta)

I know the answer is 1.6 which is 11/7 however, I am not sure why that is. The question gave me a change in theta of 6pi and two speeds. Okay, so wouldn't dividing the speeds give you the ratio of the initial speed and final speed, not the momentum? Or is the speed also the momentum?
angular momentum = moment of inertia * angular velocity, right?
Has the moment of inertia changed?
 
vysero said:
Awesome ty,

11^2 = 7^2 +2a(6pi)
a = 6pi
It helps to do the algebra before subbing in the numbers. So you are starting from: ##\omega_f^2=\omega_i^2+2\alpha\theta## and you want to solve for ##\alpha##.

vysero said:
New question, well a continuation question:

For the speeds given in problem #6 above, what is the ratio of the wheel‟s final angular
momentum to the initial angular momentum?
The whole exercise is trying to get you to think in terms of angular thingies. So momentum is "normally" ##p=mv##, and the "initial momentum" would be written ##p_i=m_iv_i## but if the mass does not change you just write ##p_i=mv_i## ... so, in angular stuff, it is: ##L=I\omega## ...

Remember to do the algebra before subbing in the numbers: - what is ##L_f/L_i##?
 
So, Lf = IWf and Li = IWi? So, IWf/IWi = Wf/Wi?
 
vysero said:
So, Lf = IWf and Li = IWi? So, IWf/IWi = Wf/Wi?
Well done!

When a physics question asks for "the ratio between a and b" they usually want a/b.
 
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