Angular Acceleration of a Wheel on a Turntable

AI Thread Summary
The discussion focuses on calculating the angular acceleration of a wheel mounted on a rotating turntable. The wheel has a constant angular velocity, while the turntable's angular velocity remains constant, affecting the direction of the wheel's angular velocity. The participant derives the angular acceleration using the formula for angular acceleration and parameterizes the wheel's angular velocity based on the turntable's motion. They realize a mistake in their calculations regarding the j term and correct it to cosine. The overall process for determining the magnitude and direction of the angular acceleration is confirmed to be on the right track.
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Homework Statement


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The axle of a wheel is mounted on supports that rest on a rotating turntable. the wheel has angular velocity ##\omega_1 = 44.0\; \frac{\textrm{rad}}{\textrm{s}}## about its axle, and the turntable has angular velocity ##\omega_2 = 35.0\; \frac{\textrm{rad}}{\textrm{s}}## about a vertical axis. What is the magnitude and direction of the angular acceleration of the wheel at the instant shown? Take the ##z## axis vertically upward and the direction of the axle at the moment shown to be the ##x## axis pointing to the right.

2x3l1z8.png


Homework Equations



##\vec{\alpha} = \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}##

The Attempt at a Solution



This problem stumped me a bit, but I think I got it now and just want to make sure. ##\vec{\omega}_2##, which is the angular velocity of the turntable, is constant because using the right hand rule it is always pointing up. ##\vec{\omega}_1## varies because the turntable causes the direction of it to change, although its magnitude is the same. Since ##\vec{\alpha} = \frac{\textrm{d}\vec{\omega}}{\textrm{d}t}##, and ##\vec{\omega}_2## is constant, the acceleration is then ##\frac{\textrm{d}\vec{\omega}_1}{\textrm{d}t}##.

The direction of ##\vec{\omega}_1## follows the circular path of ##\vec{\omega}_2##, so it can be parameterized by:

##\vec{\omega}_1 = \omega_1 \cos (\omega_2 t)\hat{i} + \omega_1 \sin(\omega_2 t)\hat{j}##

If we take ##t = 0## to be shown by the picture above, then at ##t=0## ##\vec{\omega}_1## is pointing to the left, so it has to be negative. Therefore, the parametrization should be:

##\vec{\omega}_1 = -\omega_1 \cos (\omega_2 t)\hat{i} - \omega_1 \sin(\omega_2 t)\hat{j}##

To get the acceleration, I take the derivative with respect ot time and find that:

##\vec{\alpha} = \omega_1\omega_2\sin(\omega_2 t)\hat{i} - \omega_1\omega_2\sin(\omega_2 t)\hat{j}##

I then simply have to find the magnitude and direction of this to get the answers. Was my process correct?
 
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Check the j term in your final equation.
 
haruspex said:
Check the j term in your final equation.

Whoops! should be cosine.
 
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