A merry-go-round is spinning at a rate of 3.0 revolutions per minute. Cora is sitting 1.0 m from the

5 answers
Question:

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.

Answers

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 .

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

[tex]f = \frac{4.04}{60} =0.067 rev/s[/tex]

here we know that angular frequency is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(0.067)[/tex]

[tex]\omega = 0.42 rad/s[/tex]

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

Answer with Step-by-step explanation:

Relationship between rotation and reflection:

In reflection, the pre-image gets flips across the same line.

whereas in rotation, the pre-image get spin around a point.

Rotations can be represented as two reflections i.e. the composition of reflections over two intersecting lines is always equal to rotation of that object.

In rotation, the object will get turned, this is turning some degree either clockwise or anti-clockwise.

Hence, rotation requires spinning and reflection requires flipping.


[tex]What is the relationship between a rotation and a reflection? sketch a diagram that supports your e[/tex]

Same.

Explanation:

The rotational speed of an object is given by :

[tex]\omega=\dfrac{\theta}{t}[/tex]

[tex]\theta[/tex] is the angular displacement

t is the time taken

The angular speed of a merry- go- round is 4 revolutions per minute. There are two persons Cora and Cameron. 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.

The rotational speeds of both of the children remains the same because it is independent of the distance from the center.

The rotational speed of the two children is the same. 
In fact, it is defined as
[tex]\omega = \frac{\Delta \theta}{\Delta t}[/tex]
where [tex]\Delta \theta[/tex] is the angle covered in the time [tex]\Delta t[/tex]. As it can be seen, this quantity does not depend on the distance from the centre, so the rotational speed is 4.0 revolutions per minute for both children.

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