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.

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

[tex]v=\omega r[/tex]

where

[tex]\omega[/tex] is the angular speed

r is the distance from the centre of rotation

Here let's call [tex]r_c[/tex] the distance at which Chuck is rotating, so his tangential speed is

[tex]v_c = \omega r_c[/tex]

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

[tex]r_a = 2 r_c[/tex]

So his tangential speed is

[tex]v_a = \omega r_a = \omega (2 r_c) = 2(\omega r_c) = 2 v_c[/tex]

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

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Lanuel Lanuel · Answer 2 · 2022-04-08T12:11:35+02:00

Since the merry-go-round is a rigid body, Andrea and Chuck's angular speeds will be the same and equal at all points.

What is angular speed?

Angular speed can be defined as the rate of change of angular displacement of an object with respect to time. Thus, it is a measure of how fast and quickly an object revolves (rotates) relative to another point or how its angular position changes with respect to time.

How to calculate angular speed.

Mathematically, the relationship between Andrea and Chuck's angular speeds is given by this formula:

[tex]\omega = \frac{\Delta \theta}{\Delta t}=\frac{2\pi}{T}[/tex]

Where:

[tex]\Delta \theta[/tex] is the change in angular displacement.

[tex]\Delta t[/tex] is the change in time.

T is the period.

Since the merry-go-round is a rigid body, Andrea and Chuck's angular speeds will be the same because all points on the merry-go-round cover equal angular displacement in the same amount of time.

How to calculate tangential speed.

Mathematically, tangential speed of an object in circular motion is given by this formula:

[tex]V=r\omega[/tex]

For Chucks, we have:

[tex]V_c=r_c\omega[/tex]

Note: Andrea is rotating twice as far from the centre of the circular platform as Chuck ([tex]r_a=2r_c[/tex]).

For Andrea, we have:

[tex]V_a=2V_c=2r_c\omega[/tex]

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