Question

Three children, Aricelia, Bao, and Chuck, are playing on a merry go round. Their positions on the merry go round are shown in the top view picture at right. The merry go round is rotating clockwise and is neither speeding up or slowing down

Answer 1

Each child is making same angular displacement as the other.

hence they have same angular speed. (as angular speed=angular dispalcement/time)

but they will have different speed as

linear speed=angular speed*radius

where radius=distance from the center

as all the children have different radii, they will have different linear speed

as linear speed is directly proportional to radius,

farther the child, more will be his speed.

(think of it in this way, more the radius, to cover the same angle, he/she has to cover more distance

as arc length=radius*angle, hence speed=distance covered/time will be more for the higher radius)

answers:

A. distance of chuck > distance of Bao > distance of Aricelia

==>speed of chuck >speed of Bao > speed of aricelia

B.

as explained above, they are covering same angular displacement in same time

hence all will have equal angular speed.

angular speed of chuck = angular speed of Bao =angular speed of Aricelia

C.as angular speed is equal for everybody and one revolution=2*pi radian

time taken to cover the angle =2*pi/angular speed

hence time taken for each person will be equal

time taken by chuck = time taken by Bao =time taken by Aricelia
D.

torque=moment of inertia*angular acceleration

here torque applied is constant .

to bring up to speed in as little as possible, angular acceleration should be as high as possible.

(because angular speed=angular acceleration*time

higher the angular acceleration, lesser the time required to reach a particular angular speed)

hence for a constant torque, moment of inertia should be as little as possible.

as we know, moment of inertia=mass*distance from axis^2

so lesser the distance from axis, lesser the moment of inertia for a fixed mass.

hence the three children should be as close to the axis of rotation as possible to make the merry go round come into

1 revolutions per 4 seconds in as little time as possible.