Merry-Go-Round

Classical Mechanics Level 1

A merry-go-round has a radius of about $8 \text{ m}$ . A child sitting on a horse on the outer edge sees his parents, standing still at the entrance, every $20 \text{ s}$ . If the child's horse could break free of its merry-go-round restraints and continue forward tangentially without accelerating, how fast would it be traveling in $\text{m}/\text{s}$ ?

\frac{4\pi}{5}

\frac{3\pi}{10}

\frac{\pi}{4}

\frac{\pi}{8}

2 solutions

Matt DeCross
Feb 2, 2016

The question is asking for the tangential velocity of the child, which can be found via the formula $v = r\omega$ . If the child sees his parents once every $20 \text{s}$ , the angular velocity of the merry-go-round is $2\pi / (20) \text{ rad}/\text{s} = \pi / 10 \text{ rad}/\text{s}$ . So the tangential velocity is:

$v = (8 \text{ m})(\pi / 10 \text{ rad}/\text{s}) = \frac{4\pi}{5} \text{ m}/\text{s}.$

Why isn't it the case that, if the child is moving around a merry-go-round with a radius of 8m (therefore the merry-go-round has a circumference of $64\pi$ m) every 20 seconds, that the child, released from the merry-go-round, doesn't continue forward at a rate of $64\pi$ m/20s?

You're right- I did! Dangit.

Christopher Williams - 5 years, 3 months ago

"therefore the merry-go-round has a circumference of $64 \pi$ m"

I think you combined area and circumference formulas here. Radius of $8$ m -> Circumference of $16 \pi$ m, consistent with the listed answer.

Matt DeCross - 5 years, 3 months ago

Gandoff Tan
Mar 25, 2019

$v = rω$

$v = 8(\frac {π}{10})$

$v = \boxed{\frac {4π}{5}}$

Merry-Go-Round

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