A child's top is held in place upright on a frictionless surface. The axle has a radius of r = 2.46 mm. Two strings are wrapped around the axle, and the top is set spinning by applying T = 2.90 N of constant tension to each string. If it T takes 0.770 s for the string to unwind, how much angular momentum L does the top acquire? Assume that the strings do not slip as the tension is applied. R P h kg-m? L = S Incorrect

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Chapter1: Units, Trigonometry. And Vectors
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Please help solve for angular momentum.

**Diagram Explanation**

This educational diagram depicts a child's top set up on a frictionless surface. The diagram labels several parts of the top and illustrates the forces acting upon it:

- **Axle**: The top has an axle with a radius of \( r = 2.46 \, \text{mm} \). This is where two strings are wrapped around.
- **Tension Force (T)**: Two arrows labeled \( T \) point outward, indicating the constant tension of \( T = 2.90 \, \text{N} \) applied to each string.
- **Diameter of Axle**: The distance marked \( 2r \) is the diameter of the axle, which is twice the radius.
- **Point P**: Located on the surface of the conical part of the top, representing a point where the radius \( R \) and height \( h \) intersect.
- **Angle \(\theta\)**: Indicates the angle of elevation from the base to Point P.
- **Height (\( h \))**: Represents the vertical distance from the base to the top of the conical part.

**Problem Description**

A child's top is held in place upright on a frictionless surface. The axle has a radius of \( r = 2.46 \, \text{mm} \). Two strings are wrapped around the axle, and the top is set spinning by applying \( T = 2.90 \, \text{N} \) of constant tension to each string. If it takes \( 0.770 \, \text{s} \) for the string to unwind, how much angular momentum \( L \) does the top acquire? Assume that the strings do not slip as the tension is applied.

**Angular Momentum Equation**

\[ L = \ \, \text{kg} \cdot \text{m}^2/\text{s} \]

(Incorrect answer provided in the problem)

The goal is to determine the angular momentum \( L \) acquired by the top, assuming the strings apply the tension without slipping during the unwinding process.
Transcribed Image Text:**Diagram Explanation** This educational diagram depicts a child's top set up on a frictionless surface. The diagram labels several parts of the top and illustrates the forces acting upon it: - **Axle**: The top has an axle with a radius of \( r = 2.46 \, \text{mm} \). This is where two strings are wrapped around. - **Tension Force (T)**: Two arrows labeled \( T \) point outward, indicating the constant tension of \( T = 2.90 \, \text{N} \) applied to each string. - **Diameter of Axle**: The distance marked \( 2r \) is the diameter of the axle, which is twice the radius. - **Point P**: Located on the surface of the conical part of the top, representing a point where the radius \( R \) and height \( h \) intersect. - **Angle \(\theta\)**: Indicates the angle of elevation from the base to Point P. - **Height (\( h \))**: Represents the vertical distance from the base to the top of the conical part. **Problem Description** A child's top is held in place upright on a frictionless surface. The axle has a radius of \( r = 2.46 \, \text{mm} \). Two strings are wrapped around the axle, and the top is set spinning by applying \( T = 2.90 \, \text{N} \) of constant tension to each string. If it takes \( 0.770 \, \text{s} \) for the string to unwind, how much angular momentum \( L \) does the top acquire? Assume that the strings do not slip as the tension is applied. **Angular Momentum Equation** \[ L = \ \, \text{kg} \cdot \text{m}^2/\text{s} \] (Incorrect answer provided in the problem) The goal is to determine the angular momentum \( L \) acquired by the top, assuming the strings apply the tension without slipping during the unwinding process.
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