Question

Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Chuck, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, Andrea's angular speed is which of the following?
(a) What is the relationship between their angular speeds?
(b) What is the relationship between their tangential speeds? Explain.

Answer 1

a) Their angular speeds are the same

b) Andrea's tangential speed is twice the value of Chuck's tangential speed

Explanation:

a)

The angular speed of Andrea and Chuck is the same.

Let's call $\omega$ the angular speed at which the merry-go-round is rotating. We know that the angular speed is defined as:

$\omega= \frac{2\pi}{T}$

where

$2 \pi$ is the angular displacement covered in one revolution

T is the period of revolution

The merry go round is a rigid body, so all its point cover the same angular displacement in the same time: this means that it doesn't matter where Andrea and Chuck are located along the merry-go-round, their angular speed will still be the same.

b)

For an object in circular motion, the tangential speed is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance from the centre of rotation

Here let's call $r_c$ the distance at which Chuck is rotating, so his tangential speed is

$v_c = \omega r_c$

Now we know that Andrea is rotating twice as far from the centre, so at a distance of

$r_a = 2 r_c$

So his tangential speed is

$v_a = \omega r_a = \omega (2 r_c) = 2(\omega r_c) = 2 v_c$

So, Andrea's tangential speed is twice the value of Chuck's tangential speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

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