- #1

lizzyb

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Hi. I have: A 3 kg bicycle wheel rotating at a 2484 rev/min angular velocity has its shaft supported on one side, as show in the figure. When viewing from the left (from the positive x-axes), one sees that the wheel is rotating in a clockwise manner. The distance from the center of the wheel to the pivot point is 0.6 m. The wheel is a hoop of radius 0.4 m and its shaft is horizontal.

Assume all the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9.8 m/s^2.

Find the change in the precession angle after a 1.5 s time interval.

In my book I have the equation:

[tex]d\phi = \frac{(M g h) dt}{L}[/tex]

so [tex] L = I \omega = M R^2 \omega[/tex]

therefore [tex]d\phi = \frac{(M g h) dt}{M R^2 \omega} = \frac{(g h) dt}{R^2 \omega}[/tex]

Now [tex]\omega = \frac{2484 "rev"}{"min"} \frac{1 "min"}{60 "sec"} \frac{2 \pi "rad"}{"rev"} = \frac{260.124 "rad"}{"sec"}[/tex]

and let [tex]h = 0.6, R = 0.4, g = 9.8, dt = 1.5[/tex]

Using these I came up with .211918 but the answer was wrong. Any ideas where I went wrong? Thank you.

[Edit]I tried -.211918 (the bicycle is rotating in a clockwise fashion) but that was wrong as well.

Assume all the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9.8 m/s^2.

Find the change in the precession angle after a 1.5 s time interval.

In my book I have the equation:

[tex]d\phi = \frac{(M g h) dt}{L}[/tex]

so [tex] L = I \omega = M R^2 \omega[/tex]

therefore [tex]d\phi = \frac{(M g h) dt}{M R^2 \omega} = \frac{(g h) dt}{R^2 \omega}[/tex]

Now [tex]\omega = \frac{2484 "rev"}{"min"} \frac{1 "min"}{60 "sec"} \frac{2 \pi "rad"}{"rev"} = \frac{260.124 "rad"}{"sec"}[/tex]

and let [tex]h = 0.6, R = 0.4, g = 9.8, dt = 1.5[/tex]

Using these I came up with .211918 but the answer was wrong. Any ideas where I went wrong? Thank you.

[Edit]I tried -.211918 (the bicycle is rotating in a clockwise fashion) but that was wrong as well.

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