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1. Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. Radius of tire is 50 cm. What angle did the tire move through in those 5 secs

Respuesta :

Answer:

[tex]\theta=65.18rad[/tex]

Explanation:

The angle in rotational motion is given by:

[tex]\theta=\frac{w_o+w_f}{2}t[/tex]

Recall that the angular speed is larger than regular frequency (in rpm) by a factor of [tex]2\pi[/tex], so:

[tex]\omega_f=2\pi f\\\omega_f=2\pi*250rpm\\\omega_f=1570.80 \frac{rad}{min}[/tex]

The wheel spins from rest, that means that its initial angular speed is zero([tex]\omega_o[/tex]). Finally, we have to convert the given time to minutes and replace in the first equation:

[tex]t=5s*\frac{1min}{60s}=0.083min\\\theta=\frac{\omega_f}{2}t\\\theta=\frac{1570.800\frac{rad}{min}}{2}(0.083min)\\\theta=65.18rad[/tex]