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Sagot :
Given data:
* The radius of the carnival ride is r = 3.9 m.
* The linear frequency of oscillation is f = 0.527 Hz.
Solution:
The angular frequency (angular speed) of the oscillation is,
[tex]\begin{gathered} \omega=2\pi f \\ \omega=2\pi\times0.527 \\ \omega=3.31\text{ rad/s} \end{gathered}[/tex]The frictional force acting on the ride is,
[tex]F=\mu mg[/tex]The force acting on the ride in terms of the angular speed is,
[tex]F=m\omega^2r[/tex]The frictional force acting on the ride is equal to the force acting on the ride for circular motion,
[tex]\begin{gathered} \mu mg=m\omega^2r \\ \mu g=\omega^2r \\ \mu=\frac{\omega^2r}{g} \end{gathered}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} \mu=\frac{(3.31)^2\times3.9}{9.81} \\ \mu=4.36 \end{gathered}[/tex]Thus, the coefficient of friction is 4.36.
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