Math Problem Statement

If the string does not slip while being unwound, what is the angular speed of the top after 114 cm of string has been pulled off the peg?

Solution

To solve this problem, we can use the principle of work and energy in rotational motion. Here's the detailed step-by-step solution:

Given Data

  • Moment of inertia of the top, I=0.00048kgm2I = 0.00048 \, \text{kg} \cdot \text{m}^2
  • Tension in the string, F=2.3NF = 2.3 \, \text{N}
  • Length of string pulled, s=114cm=1.14ms = 114 \, \text{cm} = 1.14 \, \text{m} (convert to meters)
  • Initial angular speed, ω0=0\omega_0 = 0 (since the top is initially at rest)

Solution Approach

  1. Calculate the Work Done by the Force (Tension) on the Top: Since the force FF is acting through a distance ss, the work done WW by the force is: W=F×sW = F \times s

  2. Relate Work Done to Rotational Kinetic Energy: The work done by the force goes into the rotational kinetic energy of the top. The rotational kinetic energy KErotKE_{\text{rot}} of a rotating object is given by: KErot=12Iω2KE_{\text{rot}} = \frac{1}{2} I \omega^2 where ω\omega is the angular speed we want to find.

    According to the work-energy principle: W=KErotW = KE_{\text{rot}} Substituting WW and KErotKE_{\text{rot}}: F×s=12Iω2F \times s = \frac{1}{2} I \omega^2

  3. Solve for ω\omega: Rearrange the equation to solve for ω\omega: ω=2×F×sI\omega = \sqrt{\frac{2 \times F \times s}{I}}

  4. Substitute the Values: ω=2×2.3×1.140.00048\omega = \sqrt{\frac{2 \times 2.3 \times 1.14}{0.00048}} Calculate each step:

    • 2×2.3×1.14=5.2442 \times 2.3 \times 1.14 = 5.244
    • 5.2440.00048=10925\frac{5.244}{0.00048} = 10925
    • ω=10925104.5rad/s\omega = \sqrt{10925} \approx 104.5 \, \text{rad/s}

Answer

The angular speed of the top after 114 cm of string has been pulled is approximately: ω104.5rad/s\omega \approx 104.5 \, \text{rad/s}

Would you like more details on any part of the solution?


Additional Related Questions:

  1. How would the solution change if the tension in the string varied with time?
  2. What is the relationship between linear and angular acceleration in this setup?
  3. How would we calculate the angular speed if the top had an initial angular velocity?
  4. What role does moment of inertia play in determining the angular speed?
  5. How would the outcome change if there were frictional forces acting on the top?

Tip:

In rotational dynamics, always consider the work-energy principle as a powerful tool to relate forces, displacements, and rotational speeds.

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Math Problem Analysis

Mathematical Concepts

Rotational Dynamics
Work-Energy Principle
Moment of Inertia

Formulas

Work Done: W = F × s
Rotational Kinetic Energy: KE_rot = 1/2 I ω^2
Angular Speed: ω = √(2Fs / I)

Theorems

Work-Energy Theorem in Rotational Motion

Suitable Grade Level

Undergraduate Physics