Question

is closer to the center. For a given elapsed time interval, which rider has greater angular displacement?
Both the girl and the boy have the same nonzero angular displacement.
Both the girl and the boy have zero angular displacement.
The boy has greater angular displacement.
The girl has greater angular displacement.
Part B

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy is closer to the center. Who has greater linear speed?

Both the girl and the boy have zero linear speed.
The girl has greater linear speed.
Both the girl and the boy have the same nonzero linear speed.
The boy has greater linear speed.

Answer 1

(a) Both the girl and the boy have the same nonzero angular displacement.

Explanation:

The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

$\theta= \omega t$

where

$\omega$ is the angular speed

t is the time interval

For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of $\omega$ is the same for the boy and the girl.

Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.

(b) The girl has greater linear speed.

Explanation:

The linear (tangential) speed of a point along the merry-go-round is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the point from the centre of the merry-go-round

In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of $\omega$ is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.