7. A small body of mass 3 kg revolves uniformly in a circle on a horizontal frictionless surface attached by a cord 0.2m. The cord is connected to a pin set in the center of the surface. If the body makes 4 revolutions in 2 seconds a. Find the centripetal force.
7. A small body of mass 3 kg revolves uniformly in a circle on a horizontal frictionless surface attached by a cord 0.2m. The cord is connected to a pin set in the center of the surface. If the body makes 4 revolutions in 2 seconds a. Find the centripetal force.
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Problem 7:
A small body of mass 3 kg revolves uniformly in a circle on a horizontal frictionless surface attached by a cord 0.2 m. The cord is connected to a pin set in the center of the surface. If the body makes 4 revolutions in 2 seconds:
**a. Find the centripetal force.**
To solve this problem, follow these steps:
1. **Calculate the velocity of the body.**
2. **Use the centripetal force formula to find the force.**
#### 1. Calculate the velocity of the body:
Given:
- Number of revolutions, \( n = 4 \)
- Time taken, \( t = 2 \) seconds
- Radius of the circle, \( r = 0.2 \) meters
First, find the angular velocity (ω):
\[ \omega = \frac{2\pi \times n}{t} = \frac{2\pi \times 4}{2} = 4\pi \, \text{rad/s} \]
Next, calculate the linear velocity (v):
\[ v = \omega \times r = 4\pi \times 0.2 = 0.8\pi \, \text{m/s} \]
#### 2. Use the centripetal force formula to find the force:
The centripetal force (F) is given by the formula:
\[ F = m \times \frac{v^2}{r} \]
Substitute the given values:
- Mass \( m = 3 \) kg
- Velocity \( v = 0.8\pi \) m/s
- Radius \( r = 0.2 \) meters
\[ F = 3 \times \frac{(0.8\pi)^2}{0.2} \]
\[ F = 3 \times \frac{0.64\pi^2}{0.2} \]
\[ F = 3 \times 3.2\pi^2 \]
\[ F = 9.6\pi^2 \, \text{N} \]
Thus, the centripetal force is approximately \( 9.6\pi^2 \, \text{N} \), which simplifies to:
\[ F \approx 94.48 \, \text{Newtons (N)} \]
Therefore, the centripetal](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffcf36abc-8605-434b-b2ea-6d36e2b820c0%2Fb3a2efb6-1863-4c38-b726-7167907b447f%2F8ss9c5.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 7:
A small body of mass 3 kg revolves uniformly in a circle on a horizontal frictionless surface attached by a cord 0.2 m. The cord is connected to a pin set in the center of the surface. If the body makes 4 revolutions in 2 seconds:
**a. Find the centripetal force.**
To solve this problem, follow these steps:
1. **Calculate the velocity of the body.**
2. **Use the centripetal force formula to find the force.**
#### 1. Calculate the velocity of the body:
Given:
- Number of revolutions, \( n = 4 \)
- Time taken, \( t = 2 \) seconds
- Radius of the circle, \( r = 0.2 \) meters
First, find the angular velocity (ω):
\[ \omega = \frac{2\pi \times n}{t} = \frac{2\pi \times 4}{2} = 4\pi \, \text{rad/s} \]
Next, calculate the linear velocity (v):
\[ v = \omega \times r = 4\pi \times 0.2 = 0.8\pi \, \text{m/s} \]
#### 2. Use the centripetal force formula to find the force:
The centripetal force (F) is given by the formula:
\[ F = m \times \frac{v^2}{r} \]
Substitute the given values:
- Mass \( m = 3 \) kg
- Velocity \( v = 0.8\pi \) m/s
- Radius \( r = 0.2 \) meters
\[ F = 3 \times \frac{(0.8\pi)^2}{0.2} \]
\[ F = 3 \times \frac{0.64\pi^2}{0.2} \]
\[ F = 3 \times 3.2\pi^2 \]
\[ F = 9.6\pi^2 \, \text{N} \]
Thus, the centripetal force is approximately \( 9.6\pi^2 \, \text{N} \), which simplifies to:
\[ F \approx 94.48 \, \text{Newtons (N)} \]
Therefore, the centripetal
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