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A merry-go-round is spinning at a rate of 3.0 revolutions per minute. Cora is sitting 1.0 m from the center of the merry-go-round and Cameron is sitting right on the edge, 2.0 m from the center.

1. What is the relationship between the rotational speed of the two children?

a. Cameron's rotational speed is double coral rotational speed.
b. Cameron's rotational speed is four times as much as coral rotational speed.
c. Cora rotational speed is double Cameron's rotational speed.
d. Cora rotational speed is the same as Cameron's rotational speed.
e. Cora rotational speed is four times as much as Cameron's rotational speed

2. What is the relationship between the tangential speed of the two children?

a. Cora tangential speed is four times as much as Cameron's rotational speed.
b. Camerons tangential speed is four times coral tangential speed.
c. Cora tangential speed is the same as Cameron's tangential speed.
d. Cora tangential speed is double Cameron's tangential speed.
e. Cameron's tangential speed is double coral tangential speed.

User Finferflu
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2 Answers

6 votes

Answer:

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Step-by-step explanation:

User Melissa Stewart
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2 votes

Answer:

Step-by-step explanation:

1 )

angular velocity of merry go round = 2π n

= 2π x 3 / 60

ω = .1 x π radian / s

This will be the rotational speed of the whole system including that of Cora and Cameron . It will not depend upon their relative position with respect to

the centre of the merry go round .

So rotational speed of Cora = Rotational speed of Cameron

option ( d ) is correct .

2 )

Tangential speed v = ω R where R is the distance from the centre of merry go round .

Tangential speed of Cora = .1 x π x 1

= .314 m /s

Tangential speed of Cameron = .1 x π x 2

= .628 m /s .

So tangential speed of Cameron is twice that of Cora .

Option ( e ) is correct .

User Steve Czetty
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