Problem 75 - University Physics with Modern Physics

Question

The Yo-yo. A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\) , connected by a light axle of radius \(b .\) A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

Step-by-Step Solution

Answer

Linear acceleration: \(a = \frac{g}{1 + \frac{R^2}{b^2}}\); Angular acceleration: \(\alpha = \frac{g}{b(1 + \frac{R^2}{b^2})}\); Tension: \(T = \frac{m g R^2}{b^2 + R^2}\).

Step 1: Determine Moment of Inertia

The yo-yo consists of two disks and an axle. The moment of inertia for each disk about its center is \( \frac{1}{2} m R^2 \). Since there are two disks, the total moment of inertia for the yo-yo is \( I = 2 \times \frac{1}{2} m R^2 = m R^2 \). Assuming the axle is lightweight, its contribution is negligible.

Step 2: Analyze Forces and Torques

Let the tension in the string be \( T \), the acceleration of the yo-yo be \( a \), and the angular acceleration be \( \alpha \). For the linear motion of the center of mass, the force equation is \( mg - T = m a \). For the rotational motion about the axle, the torque equation is \( T b = I \alpha \).

Step 3: Relate Linear and Angular Quantities

The linear acceleration \( a \) and angular acceleration \( \alpha \) are connected by the axle's radius. This gives the relation \( a = b \alpha \).

Step 4: Solve Equations Simultaneously

Substitute \( \alpha = \frac{a}{b} \) from Step 3 into the torque equation: \( T b = m R^2 \frac{a}{b} \). Simplify to get \( T = \frac{m R^2 a}{b^2} \). Use this expression for \( T \) in the linear motion equation: \( mg - \frac{m R^2 a}{b^2} = m a \). This simplifies to \( mg = m a + \frac{m R^2 a}{b^2} \).

Step 5: Solve for Linear Acceleration

Rearrange the equation \( mg = m a + \frac{m R^2 a}{b^2} \) to find \( a \). Factor to get \( g = a (1 + \frac{R^2}{b^2}) \), leading to \( a = \frac{g}{1 + \frac{R^2}{b^2}} \).

Step 6: Find Angular Acceleration

With \( a = \frac{g}{1 + \frac{R^2}{b^2}} \), use the relation \( \alpha = \frac{a}{b} \) from Step 3 to find \( \alpha = \frac{g}{b(1 + \frac{R^2}{b^2})} \).

Step 7: Calculate Tension in the String

Finally, use the earlier derived equation \( T = \frac{m R^2 a}{b^2} \) and substitute \( a = \frac{g}{1 + \frac{R^2}{b^2}} \) to find \( T = \frac{m R^2}{b^2} \times \frac{g}{1 + \frac{R^2}{b^2}} \). Simplifying, \( T = \frac{m g R^2}{b^2 + R^2} \).

Key Concepts

Moment of Inertia

Moment of inertia is a fundamental concept in rotational dynamics, playing a role similar to mass in linear dynamics. For objects that rotate, such as a yo-yo, moment of inertia determines how much torque is needed for a desired angular acceleration. In the case of a yo-yo, which consists of two uniform disks, the moment of inertia is calculated using the formula:\[ I = mR^2 \]where:

• \( I \) is the moment of inertia.
• \( m \) is the mass of one disk (since there are two disks, the equation doubles when calculated).
• \( R \) is the radius of the disks.

Considering the axle as lightweight, its contribution is often negligible. By understanding this, we can comprehend how different shapes and mass distributions affect an object's rotational resistance.

Torque

Torque is crucial in understanding rotational motion, akin to force in linear motion. It is what causes an object to rotate and is represented by the formula:\[ \tau = r \, T \]where:

• \( \tau \) is the torque.
• \( r \) is the radius at which the force is applied (in this scenario, the axle's radius \( b \)).
• \( T \) is the tension in the string providing the force.

In the yo-yo problem, the torque produced by the tension in the string causes it to unwind, converting potential energy into rotational motion.

Angular Acceleration

Angular acceleration is the rate at which an object's angular velocity changes with time. For the yo-yo problem, it is determined using the relation:\[ \alpha = \frac{a}{b} \]where:

• \( \alpha \) is the angular acceleration.
• \( a \) is the linear acceleration.
• \( b \) refers to the axle's radius.

This relationship is derived from equating the rotational and translational energies, reflecting how tightly linked linear and angular metrics are in rotational dynamics.

Linear Acceleration

Linear acceleration represents how quickly the yo-yo's velocity changes as it falls. Solving for it requires looking at forces and considering both the gravitational pull and the upward tension force:\[ a = \frac{g}{1 + \frac{R^2}{b^2}} \]where:

• \( g \) is the acceleration due to gravity.
• \( R \) and \( b \) are radii related variables as discussed before.

This equation shows a balance between gravitational and rotational resistances, making it essential to gauge the yo-yo's descent rate accurately.

Physics Problem Solving

Efficient physics problem-solving is about linking fundamental principles with mathematical expressions. For this yo-yo problem: 1. **Breakdown the System:** Understand which physical principles apply. Here, using dynamics and rotational motion principles is key. 2. **Translate Physics to Math:** Use known physics laws and relations like Newton's second law, relating force, torque, and acceleration. 3. **Solve Step-by-Step:** Identify unknowns and use algebra to find linear and angular accelerations, incorporating each into unified solutions to ensure you're considering the full system. This structured approach transforms a complex problem into manageable mathematical equations, making concepts easier to understand and solve.