Question

While her kid brother is on a wooden horse at the edge of a merry-go-round, Sheila rides her bicycle parallel to its edge. The wooden horses have a tangential speed of 6 m/s. Sheila rides at 4 m/s. The radius of the merry-go-round is 8 m.. At what time intervals does Sheila encounter her brother, if she rides in the direction of rotation of the merry-go-round?

Answer 1

Question #74462, Physics / Mechanics | Relativity

While her kid brother is on a wooden horse at the edge of a merry-go-round, Sheila rides her bicycle parallel to its edge. The wooden horses have a tangential speed of $6\mathrm{m/s}$. Sheila rides at $4\mathrm{m/s}$. The radius of the merry-go-round is $8\mathrm{m}$. At what time intervals does Sheila encounter her brother, if she rides in the direction of rotation of the merry-go-round?

Solution

The angular speed is calculated as follows.

$$\omega = \frac{\nu}{r};$$

The angular speed of Sheila is $\omega = \frac{4}{8} = 0.5 \, \mathrm{rad/s}$

The angular speed of her brother is $\omega = \frac{6}{8} = 0.75 \, \mathrm{rad/s}$

The relative angular speed is $0.75 - 0.5 = 0.25 \, \mathrm{rad/s}$

The time interval is calculated as follows.

$$t = \frac{2\pi}{0.25} = 8\pi \, \mathrm{s} \approx 25.1 \, \mathrm{s}$$

Answer: 25.1 s

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