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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.
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?