A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. If makes 5.4 rev/min, what is the velocity of the child in m/s? the merry-go-round 3.6 m/s 10.2 m/s 0.8 m/s 5.1 m/s
A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. If makes 5.4 rev/min, what is the velocity of the child in m/s? the merry-go-round 3.6 m/s 10.2 m/s 0.8 m/s 5.1 m/s
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter7: Rotational Motion And Gravitation
Section: Chapter Questions
Problem 12CQ: A child is practicing for a BMX race. His speed remains constant as he goes counterclockwise around...
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![### Circular Motion Problem
**Question:**
A child is sitting on the outer edge of a merry-go-round that is 18 meters in diameter. If the merry-go-round makes 5.4 revolutions per minute, what is the velocity of the child in meters per second (m/s)?
**Answer Choices:**
- ⬤ 3.6 m/s
- ⬤ 10.2 m/s
- ⬤ 0.8 m/s
- ⬤ 5.1 m/s
### Explanation
To solve this problem, we need to determine the linear velocity (v) of the child sitting on the edge of the merry-go-round.
1. **Calculate the Radius:**
The diameter of the merry-go-round is given as 18 meters. The radius (r) is half of the diameter.
\[
r = \frac{d}{2} = \frac{18 \text{ m}}{2} = 9 \text{ m}
\]
2. **Convert Revolutions per Minute to Revolutions per Second:**
The merry-go-round makes 5.4 revolutions per minute. We need to convert this to revolutions per second (rps).
\[
\text{Revolutions per second} = \frac{\text{Revolutions per minute}}{60} = \frac{5.4}{60} = 0.09 \text{ rps}
\]
3. **Calculate the Circumference:**
The circumference (C) of a circle is given by:
\[
C = 2 \pi r = 2 \cdot \pi \cdot 9 \text{ m} = 18 \pi \text{ m}
\]
4. **Determine the Linear Velocity:**
The linear velocity (v) is calculated by multiplying the circumference by the number of revolutions per second (rps):
\[
v = C \cdot \text{rps} = 18 \pi \text{ m} \cdot 0.09 \text{ rps} = 1.62 \pi \text{ m/s} \approx 5.1 \text{ m/s}
\]
**Correct Answer:**
- ⬤ 5.1 m/s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9bbcd5bc-9f18-4998-936a-e4dd0492e193%2F4de3c310-e7c8-4f47-96f9-fe7af4aeef6f%2Fdlsym3s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Circular Motion Problem
**Question:**
A child is sitting on the outer edge of a merry-go-round that is 18 meters in diameter. If the merry-go-round makes 5.4 revolutions per minute, what is the velocity of the child in meters per second (m/s)?
**Answer Choices:**
- ⬤ 3.6 m/s
- ⬤ 10.2 m/s
- ⬤ 0.8 m/s
- ⬤ 5.1 m/s
### Explanation
To solve this problem, we need to determine the linear velocity (v) of the child sitting on the edge of the merry-go-round.
1. **Calculate the Radius:**
The diameter of the merry-go-round is given as 18 meters. The radius (r) is half of the diameter.
\[
r = \frac{d}{2} = \frac{18 \text{ m}}{2} = 9 \text{ m}
\]
2. **Convert Revolutions per Minute to Revolutions per Second:**
The merry-go-round makes 5.4 revolutions per minute. We need to convert this to revolutions per second (rps).
\[
\text{Revolutions per second} = \frac{\text{Revolutions per minute}}{60} = \frac{5.4}{60} = 0.09 \text{ rps}
\]
3. **Calculate the Circumference:**
The circumference (C) of a circle is given by:
\[
C = 2 \pi r = 2 \cdot \pi \cdot 9 \text{ m} = 18 \pi \text{ m}
\]
4. **Determine the Linear Velocity:**
The linear velocity (v) is calculated by multiplying the circumference by the number of revolutions per second (rps):
\[
v = C \cdot \text{rps} = 18 \pi \text{ m} \cdot 0.09 \text{ rps} = 1.62 \pi \text{ m/s} \approx 5.1 \text{ m/s}
\]
**Correct Answer:**
- ⬤ 5.1 m/s
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