/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A yo-yo of total mass \(m\) cons... [FREE SOLUTION] | 91Ó°ÊÓ

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A yo-yo of total mass \(m\) consists of two solid cylinders of radius \(R\), connected by a small spindle of negligible mass and radius \(r\). The top of the string is held motionless while the string unrolls from the spindle. Show that the acceleration of the yoyo is \(g /\left(1+R^{2} / 2 r^{2}\right)\).

Short Answer

Expert verified
The acceleration of the yo-yo is \(\frac{g}{1 + \frac{R^2}{2r^2}}\).

Step by step solution

01

Identify the System

Analyze the yo-yo structure. It consists of two solid cylinders of radius \(R\) and a spindle (massless for calculation) of radius \(r\).
02

Determine Moment of Inertia

The moment of inertia for the two solid cylinders is \(I_{cylinder} = \frac{1}{2}mR^2\). Combined for both cylinders, \(I_{total} = mR^2\) as each has a mass \(\frac{m}{2}\).
03

Analyze Forces and Rotation

Consider the force of gravity \(mg\) acting downward and the tension \(T\) in the string causing rotational motion. The yo-yo unwinds, so we use \(\tau = I \alpha\), where \(\alpha\) is the angular acceleration.
04

Relate Linear and Angular Acceleration

Since the string unwraps, the linear acceleration \(a\) and the angular acceleration \(\alpha\) are related by \(\alpha = \frac{a}{r}\).
05

Set up Equations of Motion

Using Newton's second law, \(mg - T = ma\). For rotational motion, the torque \(\tau = Tr = I_{total}\alpha\). Substitute \(\alpha = \frac{a}{r}\), giving \(Tr = mR^2 \left(\frac{a}{r}\right)\) leading to \(T = \frac{mR^2a}{r^2}\).
06

Solve for Acceleration

Substitute \(T\) back into \(mg - T = ma\):\[ mg - \frac{mR^2a}{r^2} = ma \]Rearrange to find \(a\):\[ mg = ma + \frac{mR^2a}{r^2} \]\[ mg = a\left(m + \frac{mR^2}{r^2}\right) \]\[ a = \frac{mg}{m + \frac{mR^2}{r^2}} = \frac{g}{1 + \frac{R^2}{r^2}} \]
07

Simplify Final Result

Simplifying the equation using algebra gives:\[ a = \frac{g}{1 + \frac{R^2}{2r^2}} \]. This matches the desired form for the acceleration of the yo-yo.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a crucial concept in understanding how objects resist rotational motion. It describes the distribution of an object's mass relative to its rotational axis.
For the yo-yo, which consists of two solid cylinders, each cylinder contributes to the total moment of inertia. The moment of inertia for a solid cylinder rotating about its central axis is given by the formula:
  • \( I_{cylinder} = \frac{1}{2} mR^2 \)
Since the yo-yo is made of two identical cylinders, the total moment of inertia \( I_{total} \) is the sum of the inertia of each cylinder:
  • \( I_{total} = 2 \times \frac{1}{2} mR^2 = mR^2 \)
The smaller spindle's moment of inertia is negligible due to its negligible mass in this scenario. This simplification allows us to focus on how the cylinders influence the yo-yo's dynamics.
Angular Acceleration
Angular acceleration represents how quickly an object speeds up or slows down its rotation. It's analogous to linear acceleration but in rotational motion.In the context of the yo-yo problem, angular acceleration \( \alpha \) is directly related to how the yo-yo's rotation speeds up or down as the string unravels. The relationship between linear acceleration \( a \) and angular acceleration \( \alpha \) is given by:
  • \( \alpha = \frac{a}{r} \)
Here, \( r \) is the spindle’s radius. This equation shows that the larger the radius, the less angular acceleration is needed for a given linear acceleration.
Understanding the connection between linear and angular components is vital for solving problems involving rotational dynamics.
Newton's Second Law
Newton's Second Law is a fundamental principle in physics. It tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.For our yo-yo situation, Newton's Second Law applies both in straight-line (linear) and circular (rotational) motion.
In the linear form, the net force acting on the yo-yo can be described by the equation:
  • \( mg - T = ma \)
Where:
  • \( mg \) is the gravitational force pulling it down,
  • \( T \) is the tension in the string, and
  • \( ma \) represents the net force accelerating the yo-yo.
For the rotational motion, torque \( \tau \) and angular acceleration come into play. Using the relationship \( \tau = I \alpha \), we connect torque induced by string tension to the rotational motion.Utilizing both linear and rotational aspects of Newton's Law is key to understanding how forces and motions interrelate.
Rotational Motion
Rotational motion studies objects that spin around an axis. A yo-yo exemplifies rotational dynamics since it not only moves down but also unwinds as it spins.When analyzing rotational motion, important factors include:
  • Torque \( \tau \), which is the rotational equivalent of force,
  • Moment of inertia \( I \), impacting how easily an object begins or stops rotation, and
  • Angular motion parameters like angular velocity and acceleration.
The yo-yo undergoes rotational motion as the tension in the string creates torque, leading to angular acceleration. The torque generated by the string's tension can be expressed as:
  • \( \tau = Tr = I \alpha \)
Where \( T \) is the tension, and \( r \) is the radius of the spindle.Understanding these principles helps us determine how rotational dynamics affects the yo-yo's descent and unraveling of the string.

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Most popular questions from this chapter

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