College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![### Problem Statement:
**7.** A penny is placed at the outer edge of a disk (radius = 0.190 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.85 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk.
### Detailed Explanation:
To solve this problem, we'll need to apply concepts of circular motion and friction. Here's a step-by-step approach:
1. **Determine the Centripetal Force:**
The penny must experience a centripetal force to stay on the rotating disk. This force is provided by the friction between the penny and the disk's surface.
The centripetal force \( F_c \) needed to keep the penny in rotation is given by:
\[
F_c = \frac{m v^2}{r}
\]
where
- \( m \) is the mass of the penny,
- \( v \) is the tangential velocity,
- \( r \) is the radius of the disk.
2. **Calculate the Tangential Velocity:**
The tangential velocity \( v \) can be found using the relationship between velocity and the rotational period \( T \):
\[
v = \frac{2 \pi r}{T}
\]
where
- \( r \) is 0.190 m,
- \( T \) is 1.85 s.
Substituting the given values:
\[
v = \frac{2 \pi \times 0.190}{1.85} \approx 0.645 \text{ m/s}
\]
3. **Calculate the Required Centripetal Force:**
The centripetal force can now be updated as:
\[
F_c = \frac{m (0.645)^2}{0.190}
\]
4. **Determine the Minimum Coefficient of Friction:**
The frictional force \( F_{friction} \) that keeps the penny on the disk acts as the centripetal force. The frictional force is also given by:
\[
F_{friction} = \mu m g
\]
where
- \( \mu \) is the coefficient of friction,
- \( g \) is the gravitational acceleration](https://content.bartleby.com/qna-images/question/cd0a9c9c-b0c8-47bb-88dc-3ac08f167320/8b02cc27-768d-4f4f-b1c3-b9b6e1fef2d1/udo0si7_processed.jpeg)
Transcribed Image Text:### Problem Statement:
**7.** A penny is placed at the outer edge of a disk (radius = 0.190 m) that rotates about an axis perpendicular to the plane of the disk at its center. The period of the rotation is 1.85 s. Find the minimum coefficient of friction necessary to allow the penny to rotate along with the disk.
### Detailed Explanation:
To solve this problem, we'll need to apply concepts of circular motion and friction. Here's a step-by-step approach:
1. **Determine the Centripetal Force:**
The penny must experience a centripetal force to stay on the rotating disk. This force is provided by the friction between the penny and the disk's surface.
The centripetal force \( F_c \) needed to keep the penny in rotation is given by:
\[
F_c = \frac{m v^2}{r}
\]
where
- \( m \) is the mass of the penny,
- \( v \) is the tangential velocity,
- \( r \) is the radius of the disk.
2. **Calculate the Tangential Velocity:**
The tangential velocity \( v \) can be found using the relationship between velocity and the rotational period \( T \):
\[
v = \frac{2 \pi r}{T}
\]
where
- \( r \) is 0.190 m,
- \( T \) is 1.85 s.
Substituting the given values:
\[
v = \frac{2 \pi \times 0.190}{1.85} \approx 0.645 \text{ m/s}
\]
3. **Calculate the Required Centripetal Force:**
The centripetal force can now be updated as:
\[
F_c = \frac{m (0.645)^2}{0.190}
\]
4. **Determine the Minimum Coefficient of Friction:**
The frictional force \( F_{friction} \) that keeps the penny on the disk acts as the centripetal force. The frictional force is also given by:
\[
F_{friction} = \mu m g
\]
where
- \( \mu \) is the coefficient of friction,
- \( g \) is the gravitational acceleration
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