3. A telephone pole has been knocked over by the wind so that it makes an angle of 0 with the vertical. The wind has stopped blowing and the pole is to be cut down as shown. nail cut Once the cut is across most of the thickness of the pole, the pole begins to tip over. As the pole tips, the bottom of the pole stays attached to the base by the remaining part of the pole that was not cut (but the torque from the base on the tipping pole is negligible). There is a nail (of negligible mass) on the pole located a distance x. The length of the pole from the cut to the top is L. Remember that the moment of inertia of a stick of mass m and length L about its end is ml?.
3. A telephone pole has been knocked over by the wind so that it makes an angle of 0 with the vertical. The wind has stopped blowing and the pole is to be cut down as shown. nail cut Once the cut is across most of the thickness of the pole, the pole begins to tip over. As the pole tips, the bottom of the pole stays attached to the base by the remaining part of the pole that was not cut (but the torque from the base on the tipping pole is negligible). There is a nail (of negligible mass) on the pole located a distance x. The length of the pole from the cut to the top is L. Remember that the moment of inertia of a stick of mass m and length L about its end is ml?.
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![**Problem Statement:**
A telephone pole has been knocked over by the wind so that it makes an angle of \( \theta \) with the vertical. The wind has stopped blowing, and the pole is to be cut down as shown.
**Diagram Explanation:**
The diagram illustrates a slanted pole with the following elements:
- A dotted line indicates the vertical axis, making an angle \( \theta \) with the slanted pole.
- A nail is positioned on the pole, located a distance \( x \) from the cut.
- \( L \) represents the length of the remaining pole from the cut to the top.
- The label "cut" marks the lower end of the pole.
**Further Explanation:**
Once the cut is across most of the thickness of the pole, it begins to tip over. As it tips, the bottom of the pole stays attached to the base due to the remaining uncut portion. The torque from the base on the tipping pole is negligible. There is a nail (assumed to have negligible mass) positioned on the pole at a distance \( x \).
Remember that the moment of inertia of a stick of mass \( m \) and length \( L \) about its end is given by:
\[
\frac{1}{3}mL^2
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f7a11c6-35df-4daf-855c-043e8f69fb75%2Fa88fa8b6-6004-45f2-be78-e50556e4fa22%2Fvtfrgze_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A telephone pole has been knocked over by the wind so that it makes an angle of \( \theta \) with the vertical. The wind has stopped blowing, and the pole is to be cut down as shown.
**Diagram Explanation:**
The diagram illustrates a slanted pole with the following elements:
- A dotted line indicates the vertical axis, making an angle \( \theta \) with the slanted pole.
- A nail is positioned on the pole, located a distance \( x \) from the cut.
- \( L \) represents the length of the remaining pole from the cut to the top.
- The label "cut" marks the lower end of the pole.
**Further Explanation:**
Once the cut is across most of the thickness of the pole, it begins to tip over. As it tips, the bottom of the pole stays attached to the base due to the remaining uncut portion. The torque from the base on the tipping pole is negligible. There is a nail (assumed to have negligible mass) positioned on the pole at a distance \( x \).
Remember that the moment of inertia of a stick of mass \( m \) and length \( L \) about its end is given by:
\[
\frac{1}{3}mL^2
\]
![### Angular Motion of the Nail
a. **Angular Acceleration as a Function of \( \theta \)**
To find the angular acceleration of the nail as a function of \( \theta \), use a coordinate system with the \( \hat{z} \) axis pointing out of the page. The angular acceleration \( \vec{\alpha} \) is then given by:
\[
\vec{\alpha} = -\frac{3g}{2L} \sin \theta \, \hat{z}
\]
b. **Angle for Tangential Acceleration Equal to \( g \)**
Determine the angle at which the magnitude of the tangential acceleration of the nail equals \( g \). The solution gives:
\[
\sin \theta = \frac{2L}{3x}
\]
c. **Magnitude of Centripetal Acceleration**
If the angular velocity of the pole at any angle is denoted \( |\vec{\omega}| \), find the magnitude of the centripetal acceleration of the nail at that angle.
d. **Total Acceleration as a Function of \( \theta \) and \( |\vec{\omega}| \)**
Derive an expression for the magnitude of the total acceleration of the nail, combining both angular and centripetal components, as a function of \( \theta \) and \( |\vec{\omega}| \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f7a11c6-35df-4daf-855c-043e8f69fb75%2Fa88fa8b6-6004-45f2-be78-e50556e4fa22%2Faxufoei_processed.png&w=3840&q=75)
Transcribed Image Text:### Angular Motion of the Nail
a. **Angular Acceleration as a Function of \( \theta \)**
To find the angular acceleration of the nail as a function of \( \theta \), use a coordinate system with the \( \hat{z} \) axis pointing out of the page. The angular acceleration \( \vec{\alpha} \) is then given by:
\[
\vec{\alpha} = -\frac{3g}{2L} \sin \theta \, \hat{z}
\]
b. **Angle for Tangential Acceleration Equal to \( g \)**
Determine the angle at which the magnitude of the tangential acceleration of the nail equals \( g \). The solution gives:
\[
\sin \theta = \frac{2L}{3x}
\]
c. **Magnitude of Centripetal Acceleration**
If the angular velocity of the pole at any angle is denoted \( |\vec{\omega}| \), find the magnitude of the centripetal acceleration of the nail at that angle.
d. **Total Acceleration as a Function of \( \theta \) and \( |\vec{\omega}| \)**
Derive an expression for the magnitude of the total acceleration of the nail, combining both angular and centripetal components, as a function of \( \theta \) and \( |\vec{\omega}| \).
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