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Magazine Chapter 10: Problem 67
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.
Short Answer
Expert verified
Linear acceleration: \( a = \frac{2b^2g}{b^2 + R^2} \), angular acceleration: \( \alpha = \frac{2b g}{b^2 + R^2} \), string tension: \( T = \frac{2mR^2 g}{b^2 + R^2} \).
Step by step solution
01
Understand the Problem
The yo-yo is composed of two disks connected by an axle, and the problem is to find the linear acceleration of its center of mass, its angular acceleration, and the tension in the string as it unwinds.
02
Identify the Forces and Torques
The force acting on the yo-yo is gravity, with a force of magnitude \( 2mg \) since there are two disks. Tension \( T \) acts upward along the string. The torque \( \tau \) about the axle due to tension is \( T \times b \), where \( b \) is the axle's radius.
03
Apply Newton's Second Law for Translation
For linear motion, use \( ma = 2mg - T \), where \( a \) is the linear acceleration of the yo-yo. The net force equals the mass times acceleration of the center of mass.
04
Apply Newton's Second Law for Rotation
For rotation, the equation is \( I \alpha = T \times b \), where \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration. Since the disks rotate together, the total moment of inertia \( I \) is \( 2 \times \frac{1}{2}mR^2 = mR^2 \).
05
Relate Linear and Angular Quantities
Relate the linear acceleration \( a \) to angular acceleration \( \alpha \) using the axle radius: \( a = \alpha b \).
06
Solve for Angular Acceleration
Substitute \( \alpha = \frac{a}{b} \) into the rotational equation: \( mR^2 \frac{a}{b} = T \times b \). Solve for \( T \) to get \( T = \frac{mR^2 a}{b^2} \).
07
Solve for Linear Acceleration
Substitute the expression for \( T \) into the linear equation: \( ma = 2mg - \frac{mR^2 a}{b^2} \). Solve for \( a \) to obtain \( a = \frac{2b^2g}{b^2 + R^2} \).
08
Find Angular Acceleration
With \( a = \frac{2b^2g}{b^2 + R^2} \), find \( \alpha = \frac{a}{b} = \frac{2b g}{b^2 + R^2} \).
09
Calculate Tension in the String
Use the formula derived earlier: \( T = \frac{mR^2 a}{b^2} = \frac{2mR^2 g}{b^2 + R^2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Torque
When a yo-yo is released, it experiences various forces and movements. Torque is a crucial concept to understand in this context. Torque (\( \tau \)) is the rotational equivalent of force. It determines how effectively a force causes an object to rotate about an axis. For the yo-yo problem, the torque is produced by the tension in the string that winds around the axle's radius.
- The formula for torque is \( \tau = T \times b \), where \( T \) is the tension in the string and \( b \) is the axle's radius.
- This torque results in angular acceleration, making the yo-yo spin as it descends.
- Understanding torque allows us to analyze how forces cause rotational motion.
Newton's Second Law
Newton's Second Law plays a pivotal role in understanding both linear and rotational dynamics of the yo-yo. This law can be applied to determine the linear motion and tension in the string as the yo-yo descends.**For Linear Motion:**
- Newton's Second Law states that the net force (\( F_{net} \)) acting on an object is equal to the object鈥檚 mass (\( m \)) multiplied by its linear acceleration (\( a \)). In equation form, it is: \( F_{net} = ma \).
- In the case of the yo-yo, the gravitational force pulling it downward is \( 2mg \) (since there are two disks, each contributing \( mg \)), and the tension \( T \) acts upward through the string.
- The resulting equation becomes: \( ma = 2mg - T \)
- Newton's Second Law for rotation can be expressed as \( \tau = I\alpha \), where \( I \) is the moment of inertia and \( \alpha \) is the angular acceleration.
- It shows how much torque is needed for an object to overcome its inertial reluctance to change rotational speed.
Angular Acceleration
Angular acceleration is how quickly an object speeds up or slows down its rotational motion. For the yo-yo problem, this is directly influenced by the torque applied by the string tension.**Key Points about Angular Acceleration:**
- Denoted by \( \alpha \), angular acceleration is measured in radians per second squared (\( \text{rad/s}^2 \)).
- It鈥檚 related to linear acceleration (\( a \)) through the radius of the axle: \( a = \alpha b \).
- The expression for \( \alpha \) in the yo-yo context derived from the problem is \( \alpha = \frac{2bg}{b^2 + R^2} \).
- Knowing the relationship between \( a \) and \( \alpha \)
- Using the given physical properties like radii (\( b \) and \( R \))
Moment of Inertia
The moment of inertia (\( I \)), often referred to as the rotational inertia, is a measure of an object鈥檚 resistance to changes in its rotational state. For the yo-yo made of two disks, this concept is quite significant in calculating the rotation as it falls.**Things to Know about Moment of Inertia:**
- For a single disk, the moment of inertia about its center is given by \( \frac{1}{2}mR^2 \).
- Since the yo-yo consists of two disks, the total moment of inertia is \( I = 2 \times \frac{1}{2}mR^2 = mR^2 \).
- This value is crucial in calculating angular acceleration using \( \tau = I \alpha \).
