/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A 5.0-m-diameter merry-go-round ... [FREE SOLUTION] | 91Ó°ÊÓ

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A 5.0-m-diameter merry-go-round is turning with a 4.0 s period. What is the speed of a child on the rim?

Short Answer

Expert verified
The speed of a child on the merry-go-round's rim is 3.9 m/s.

Step by step solution

01

Determine the radius of the merry-go-round

Given, the diameter of the merry-go-round is 5.0 m. Therefore, the radius \(r\) can be obtained by dividing the diameter by 2, which gives \(r = 5.0 m /2 = 2.5 m\).
02

Calculate the circumference of the merry-go-round

The distance covered by the child in one complete rotation around the merry-go-round is its circumference. The formula to calculate the circumference \(C\) of a circle is \(C = 2Ï€r\). Substituting \(r = 2.5 m\) into the equation, we get \(C = 2Ï€*2.5 m = 15.7 m\).
03

Determine the total time of one rotation

The given period is the time it takes for one complete rotation of the merry-go-round, which is 4.0 s.
04

Calculate the speed of the child

The speed \(v\) can be determined by dividing the total distance (which in this case is the circumference) by the total time (which is the period). Therefore, using the formula \(v = distance / time\), substituting the values obtained gives \(v = 15.7 m / 4.0 s = 3.9 m/s\). The speed of the child on the rim is therefore \(3.9 m/s\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Motion
Rotational motion describes the movement of a body in a circular path around a central point or axis. This is an integral concept when analyzing motions like that of a merry-go-round. In such a scenario, each point on the merry-go-round follows a circular trajectory, maintaining a constant distance from the center.
A critical aspect of rotational motion is that all points on the object move through the same angle in the same amount of time, but not at the same linear speed. The linear speed is highest at the outer edge and decreases as you move closer to the center, which is why the speed of the child on the rim is different from that of a point closer to the center of the merry-go-round.
Circumference of a Circle
The circumference of a circle is of particular importance when solving problems involving the motion of objects in circular paths. It is the total distance around the circle, equivalent to the perimeter of a polygon. In the given merry-go-round problem, knowing the circumference allows us to ascertain the distance travelled by a child sitting on the edge.
The formula for the circumference is given by \( C = 2\pi r \), with \( r \) being the radius of the circle. If a child were to hold a paintbrush to the merry-go-round, the length of the paint line created after one full rotation would be equivalent to the circumference.
Angular Velocity
Angular velocity describes the rate of change of angular displacement and is a measure of the speed of rotation. It refers to how fast an object rotates or revolves relative to another point, like how quickly the merry-go-round is spinning. Although the term angular velocity does not appear explicitly in the problem, it's inherently connected with the concept of rotational motion.
The relationship between angular velocity (usually denoted as \( \omega \) ) and the linear velocity (\( v \) ) is crucial. For example, on a merry-go-round, while the angular velocity is the same for every point, the linear velocity at the rim, where the child is located, is the highest due to the greater radius. The relation is typically expressed as \( v = ω r \).
Periodic Motion
Periodic motion is a type of motion that repeats itself at regular time intervals, such as the rotation of a merry-go-round. The period (\( T \) ) is the time it takes to complete one cycle of motion. It's a crucial concept given in the problem as 4.0 seconds, which denotes the time for the merry-go-round to make one full rotation.
Understanding the pattern of periodic motion can help us predict future positions of an object. This predictability is practical and extends beyond merry-go-rounds to other phenomena such as the orbits of planets, making it a fundamental element in physics.

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