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Chapter 10: Problem 98

A yo-yo-shaped device mounted on a horizontal frictionless axis is used to lift a 30 \(\mathrm{kg}\) box as shown in Fig. \(10-56\). The outer radius \(R\) of the device is \(0.50 \mathrm{~m}\), and the radius \(r\) of the hub is \(0.20 \mathrm{~m}\). When a constant horizontal force \(\vec{F}_{\text {app }}\) of magnitude \(140 \mathrm{~N}\) is applied to a rope wrapped around the outside of the device, the box, which is suspended from a rope wrapped around the hub, has an upward acceleration of magnitude \(0.80 \mathrm{~m} / \mathrm{s}^{2} .\) What is the rotational inertia of the device about its axis of rotation?

Short Answer

Expert verified
The rotational inertia of the device is \(1.55\, \mathrm{kg \cdot m^2}\).

Step by step solution

01

Identify Given Values

We have the mass of the box, \( m = 30 \mathrm{~kg} \). The outer radius \( R = 0.50 \mathrm{~m} \) and the hub radius \( r = 0.20 \mathrm{~m} \). The applied force \( F_{\text{app}} = 140 \mathrm{~N} \), and the upward acceleration of the box \( a = 0.80 \mathrm{~m/s^2} \).
02

Write Force Equations for the System

Apply Newton's second law for the box: \( T - mg = ma \), where \( T \) is the tension in the rope. Rearranging gives \( T = m(g + a) \). Using \( g = 9.8 \mathrm{~m/s^2} \), we substitute to get \( T = 30(9.8 + 0.80) = 319 \mathrm{~N} \).
03

Torque on the Yo-Yo Device

The torque due to the applied force is \( \tau_{\text{app}} = F_{\text{app}} \times R = 140 \times 0.50 = 70 \mathrm{~Nm} \). Similarly, the torque due to tension in the rope on the hub is \( \tau = T \times r = 319 \times 0.20 = 63.8 \mathrm{~Nm} \).
04

Calculate Net Torque

The net torque \( \tau_{\text{net}} = \tau_{\text{app}} - \tau = 70 - 63.8 = 6.2 \mathrm{~Nm} \).
05

Newton's Second Law for Rotation

Using \( \tau_{\text{net}} = I \alpha \), where \( I \) is the rotational inertia and \( \alpha \) is the angular acceleration. From \( a = \alpha \times r \), we have \( \alpha = \frac{a}{r} = \frac{0.80}{0.20} = 4 \mathrm{~rad/s^2} \).
06

Solve for Rotational Inertia

Rearranging the rotational form of Newton's second law gives \( I = \frac{\tau_{\text{net}}}{\alpha} = \frac{6.2}{4} = 1.55 \mathrm{~kg \cdot m^2} \).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle that describes how the velocity of an object changes when it is subjected to an external force. In simple terms, the law is often expressed as the equation:
  • F = ma
where
  • F is the net force acting on an object,
  • m is the mass of the object,
  • a is the acceleration of the object.

In the context of the yo-yo system, Newton's Second Law helps us determine the tension in the rope lifting the box. By substituting the known values (mass of the box, gravitational acceleration, and box's upward acceleration), we calculate the tension needed to achieve the desired motion. This step is crucial for later calculating the net torque applied to the yo-yo.
Torque
Torque is the rotational equivalent of linear force. It measures how much a force causes an object to rotate around an axis. Torque is represented by the formula:
  • \( \tau = F \times r \)
where
  • \( \tau \) is the torque,
  • F is the force applied,
  • r is the distance from the axis of rotation to the point where the force is applied.

In the yo-yo exercise, two torques affect the system:
  • the torque from the applied horizontal force acting on the outer radius,
  • the torque from the tension in the rope on the hub's radius.
By calculating these torques, we determine how they influence the rotational motion of the yo-yo. The difference, or net torque, is what enables us to find the angular acceleration of the device.
Angular Acceleration
Angular acceleration is the rate at which an object's angular velocity changes over time. It is connected to linear acceleration through the radius of the motion path. The relationship between linear acceleration (\( a \)) and angular acceleration (\( \alpha \)) is given by:
  • \( a = \alpha \times r \)
where
  • \( r \) is the radius of rotation.

In our yo-yo example, the linear acceleration of the box translates into angular acceleration of the yo-yo through the hub's radius. By dividing the linear acceleration by the hub's radius, we find the yo-yo's angular acceleration. This parameter is essential for understanding how quickly the yo-yo will spin around its horizontal axis.
Yo-Yo System
The yo-yo system in this exercise is a clever application of rotational mechanics principles. Imagine a traditional yo-yo, but much larger, and used to lift a load. Here's how the key components interact:
  • Outer Radius (\( R \)): The diameter where the force is applied, causing the entire system to rotate.
  • Hub Radius (\( r \)): Where the rope connected to the load is wound, translating linear force into rotational movement.
  • Rotational Inertia (\( I \)): This is the resistance of the yo-yo to changes in its rotational motion, akin to mass in linear motion.

These features work together to lift the box smoothly. When a force is applied to the yo-yo, it creates rotational motion, transferring energy from the horizontal motion into lifting energy. By understanding the yo-yo's design, students can calculate the precise rotational inertia, demonstrating how key physics principles give insights into real-world applications.

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Most popular questions from this chapter

A uniform cylinder of radius \(10 \mathrm{~cm}\) and mass \(20 \mathrm{~kg}\) is mounted so as to rotate freely about a horizontal axis that is parallel to and \(5.0 \mathrm{~cm}\) from the central longitudinal axis of the cylinder. (a) What is the rotational inertia of the cylinder about the axis of rotation? (b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

An astronaut is being tested in a centrifuge. The centrifuge has a radius of \(10 \mathrm{~m}\) and, in starting, rotates according to \(\theta=0.30 t^{2}\), where \(t\) is in seconds and \(\theta\) is in radians. When \(t=5.0 \mathrm{~s}\), what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?

The angular position of a point on a rotating wheel is given by \(\theta=2.0+4.0 t^{2}+2.0 t^{3}\), where \(\theta\) is in radians and \(t\) is in seconds. At \(t=0\), what are (a) the point's angular position and (b) its angular velocity? (c) What is its angular velocity at \(t=4.0 \mathrm{~s}\) ? (d) Calculate its angular acceleration at \(t=2.0 \mathrm{~s}\). (e) Is its angular acceleration constant?

A pulley, with a rotational inertia of \(1.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its axle and a radius of \(10 \mathrm{~cm}\), is acted on by a force applied tangentially at its rim. The force magnitude varies in time as \(F=0.50 t+0.30 t^{2}\), with \(F\) in newtons and \(t\) in seconds. The pulley is initially at rest. At \(t=3.0 \mathrm{~s}\) what are its (a) angular acceleration and (b) angular speed?

Four particles, each of mass, \(0.20 \mathrm{~kg}\), are placed at the vertices of a square with sides of length \(0.50 \mathrm{~m}\). The particles are connected by rods of negligible mass. This rigid body can rotate in a vertical plane about a horizontal axis \(A\) that passes through one of the particles. The body is released from rest with rod \(A B\) horizontal (Fig. \(10-61\) ). (a) What is the rotational inertia of the body about axis \(A ?\) (b) What is the angular speed of the body about axis \(A\) when rod \(A B\) swings through the vertical position?
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