Math Problem Statement

A uniform circular disk with radius R and mass M is wound with a string and hung. Find the following quantities when the disk is released: 1. Tension in the string 2. Acceleration of the disk's center of mass 3. Angular acceleration of the disk

Solution

The problem appears to be from a physics textbook and involves rotational motion, specifically a uniform disk of radius RR and mass MM that is hanging by a string wound around its edge. The goal is to solve for:

  1. The tension in the string (TT),
  2. The acceleration of the center of mass (acma_{cm}),
  3. The angular acceleration (α\alpha) of the disk.

Explanation:

Concepts:

  • Torque and Rotational Dynamics: Torque (τ\tau) is related to angular acceleration (α\alpha) by: τ=Iα\tau = I \alpha where II is the moment of inertia of the disk about its center. For a uniform disk: I=12MR2I = \frac{1}{2} M R^2

  • Force Analysis: The forces acting on the disk are:

    • Gravitational force: Fg=MgF_g = Mg,
    • Tension in the string: TT.
  • The linear acceleration acma_{cm} of the center of mass is related to the angular acceleration by: acm=Rαa_{cm} = R \alpha

Equations:

  1. Force Equation (translation of the center of mass): MgT=MacmMg - T = Ma_{cm}

  2. Torque Equation (rotation about the center): τ=TR=Iα\tau = T R = I \alpha Substituting I=12MR2I = \frac{1}{2} M R^2: TR=12MR2αT R = \frac{1}{2} M R^2 \alpha

    Simplify to: T=12MRαT = \frac{1}{2} M R \alpha

  3. Relation between acma_{cm} and α\alpha: acm=Rα    α=acmRa_{cm} = R \alpha \implies \alpha = \frac{a_{cm}}{R}


Solving the System:

  1. From the torque equation: T=12MacmT = \frac{1}{2} M a_{cm}

  2. Substitute T=12MacmT = \frac{1}{2} M a_{cm} into the force equation: Mg12Macm=MacmMg - \frac{1}{2} M a_{cm} = M a_{cm}

    Simplify: Mg=32MacmMg = \frac{3}{2} M a_{cm}

    Solve for acma_{cm}: acm=23ga_{cm} = \frac{2}{3} g

  3. Angular acceleration α\alpha: α=acmR=23gR=2g3R\alpha = \frac{a_{cm}}{R} = \frac{\frac{2}{3} g}{R} = \frac{2g}{3R}

  4. Tension TT: T=12Macm=12M23g=13MgT = \frac{1}{2} M a_{cm} = \frac{1}{2} M \cdot \frac{2}{3} g = \frac{1}{3} Mg


Final Answers:

  1. Tension in the string: T=13MgT = \frac{1}{3} Mg,
  2. Acceleration of the center of mass: acm=23ga_{cm} = \frac{2}{3} g,
  3. Angular acceleration: α=2g3R\alpha = \frac{2g}{3R}.

Would you like a deeper explanation or derivation of any specific step? Let me know!


Related Questions:

  1. What would change if the object were a ring instead of a disk?
  2. How does the moment of inertia influence the acceleration?
  3. Can you derive these equations for a non-uniform object?
  4. What happens if the string has mass? How does that affect the tension?
  5. How is energy conserved in this system?

Tip:

Always check if the rotational and linear dynamics are properly coupled through acm=Rαa_{cm} = R \alpha. It ensures consistency in solving such problems.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Rotational Motion
Torque
Moment of Inertia
Newton's Laws of Motion

Formulas

Torque: τ = I * α
Moment of Inertia for a disk: I = 1/2 * M * R^2
Force equation: Mg - T = M * a_cm
Relation between linear and angular motion: a_cm = R * α

Theorems

Newton's Second Law for Rotation: τ = I * α
Newton's Second Law for Translation: F_net = ma

Suitable Grade Level

Grades 11-12 or Introductory College Physics