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physicsss
Oct19-04, 10:33 PM
The axle of a wheel is mounted on supports that rest on a rotating turntable as shown in Fig. 10-50. The wheel has angular velocity w1 = 50.0 rad/s about its axle, and the turntable has angular velocity of w=35.0 rad/s about a vertical axis. (Note arrows showing these motions in the figure.)
here's the picture: http://www.geocities.com/sinceury/10-50.gif
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.
The angular velocity of the wheel is
\omega_{wheel} \left(\cos \omega_{table} t \hat i - \sin \omega_{table} t \hat j \right) + \omega_{"table"} \hat k
from which you can find the angular acceleration by differentiating with respect to time. (I didn't check the signs carefully so check them - this will give you a start!)
physicsss
Oct20-04, 11:11 PM
I don't understand...
If the table were not turning the angular velocity of the wheel would be a vector with only x and y components. What I did was to take that angular velocity and rotated it as a function of time about the z axis. That's the first part of the expression I wrote. In fact, I chose the rotation rate to be that of the table upon which the wheel sits. (The \omega's are angular speeds - i.e. not vectors!.)
Imparting a rotation of the wheel about the z axis provides an additional component of the angular velocity of the wheel. That is the second term I wrote.
physicsss
Oct21-04, 12:59 AM
The angular acceleration acceleration is 50(cos35-sin35)...?
No. The expression I wrote is the angular velocity. To find the angular acceleration you will need to differentiate with respect to time. Don't forget the angular acceleration is a vector meaning it has both direction and magnitude.
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