/*! 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 141 A disk of radius \(r\) rolls to ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A disk of radius \(r\) rolls to the right with a constant velocity \(\mathrm{v}\). Denoting by \(P\) the point of the rim in contact with the ground at \(t=0\), derive expressions for the horizontal and vertical components of the velocity of \(P\) at any time \(t\)

Short Answer

Expert verified
The velocities are \(v_x = v(1 - \cos(\frac{vt}{r}))\) and \(v_y = v \sin(\frac{vt}{r})\).

Step by step solution

01

Understand the Problem

A disk of radius \(r\) rolls to the right with a constant velocity \(v\). At time \(t = 0\), point \(P\) is at the contact point on the ground. We need to find the expressions for the horizontal and vertical components of velocity of point \(P\) after time \(t\).
02

Set the Reference for the Rolling Disk

As the disk rolls, point \(P\) moves in a circular path with radius \(r\). The disk's center moves horizontally at a constant velocity \(v\), and point \(P\) rotates around the center of the disk.
03

Horizontal Velocity Component

To find the horizontal component of velocity of point \(P\), consider its relation to the rolling velocity and the rotational motion. The total horizontal velocity is the sum of the translational velocity \(v\) and the horizontal component due to rotation, which is \(-v\cos(\omega t)\). Thus, the horizontal component of velocity is:\[ v_x = v - v \cos(\omega t) = v(1 - \cos(\omega t)) \]
04

Vertical Velocity Component

Now, calculate the vertical component of velocity due to rotation. Since the point rotates in a circle, the vertical component is given by the rotational velocity term \(v \sin(\omega t)\). Therefore, the vertical component of velocity is:\[ v_y = v \sin(\omega t) \]
05

Define Angular Speed \(\omega\)

The angular speed \(\omega\) is related to the linear speed \(v\), because as the disk rolls without slipping, \(v = r\omega\). Therefore, \(\omega = \frac{v}{r}\).
06

Final Expressions for Velocity Components

Using the value of \(\omega\), substitute back into the expressions for velocity components:Horizontal component: \[ v_x = v(1 - \cos(\frac{vt}{r})) \] Vertical component:\[ v_y = v \sin(\frac{vt}{r}) \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Velocity Components
When studying the kinematics of rolling motion, one of the key ideas is how the velocity of a point on the surface of a rolling object, like a disk, can be broken into different components. These components are the different parts of velocity in distinct directions. For instance, the velocity of point \(P\) on our rolling disk can be divided into horizontal and vertical components:

  • The horizontal component is influenced by the disk's translational motion and its rotation around its center.
  • The vertical component comes solely from the disk rotating about its center.
Such analysis helps in understanding the movement fully, and for point \(P\), the horizontal component of velocity is \(v_x = v(1 - \cos(\frac{vt}{r}))\), while the vertical component is \(v_y = v \sin(\frac{vt}{r})\). These expressions denote how the velocities change over time as the disk continues to roll.
Exploring Rotational Motion
Rotational motion refers to the movement of an object around its axis. When considering the kinematics of a rolling disk, rotational motion plays a vital role. Each point on the rim of the disk follows a circular path around the central axis.

  • In our exercise, point \(P\) moves not just horizontally but also rotates around the center.
  • The rotational motion affects both the horizontal and vertical velocity components of the point.
The blending of the disk's linear travel and its rotational spin creates the complex motion we observe in rolling objects. This understanding is crucial, as it lays down the foundation for calculating other parameters like angular speed and how they influence the movement of point \(P\).
Decoding Angular Speed
Angular speed, often represented by \(\omega\), is the rate at which an object rotates. For a rolling disk, this factor is directly tied to its linear speed \(v\) because of the no-slip condition. This means the disk rolls without sliding, creating a unique relationship between these two aspects of motion.

- For our disk, \(\omega\) is calculated by the formula \(\omega = \frac{v}{r}\), where \(r\) is the radius of the disk.- A critical part of the solution for velocity components, \(\omega\) determines the circular rate of movement.

Through this conversion, it is possible to understand how fast the disk spins in relation to how rapidly it moves forward. This piece of information enables creating accurate velocity expressions for any point on the disk, such as point \(P\).
Translational Velocity Explained
The concept of translational velocity refers to the linear motion of the entire disk as it rolls. This velocity is the mean speed at which the center of the disk moves in a straight line. For our disk rolling to the right, the translational velocity is essentially the constant velocity \(v\) as given in the problem statement.

  • This motion is straightforward and unidirectional, without changes in direction or speed.
  • In the context of kinematics, translational velocity is complemented by rotational motion to account for the disk's comprehensive motion.
While translational velocity concerns the movement of the disk's center, it crucially influences the overall velocity of points on the disk surface. This dual dynamic of translational and rotational motion provides a full picture of how each part of the disk, including point \(P\), behaves as the motion continues.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 3 -in-radius drum is rigidly attached to a 5 -in.-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that at the instant shown end \(D\) of the cord has a velocity of 8 in \(/ \mathrm{s}\) and an acceleration of \(30 \mathrm{in} / \mathrm{s}^{2}\), both directed to the left, determine the accelerations of points \(A, B,\) and \(C\) of the drums.

A series of small machine components being moved by a conveyor belt pass over a 120 -mm-radius idler pulley. At the instant shown, the angular velocity of the idler pulley is 4 rad/s clockwise. Determine the angular acceleration of the pulley for which the magnitude of the total acceleration of the machine component at \(B\) is \(2400 \mathrm{mm} / \mathrm{s}^{2}\).

$$ \begin{array}{l}{\text { Rod } A B \text { is welded to the } 0.3 \text { -m-radius plate that rotates at the }} \\ {\text { constant rate } \omega_{1}=6 \text { rad/s. Knowing that collar } D \text { moves toward }} \\\ {\text { end } B \text { of the rod at a constant speed } u=1.3 \mathrm{m} / \mathrm{s} \text { , determine, for }} \\ {\text { the position shown, (a) the velocity of } D,(b) \text { the acceleration of } D .}\end{array} $$

A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of \(3600 \mathrm{rpm}\). When the power is turned on, the unit reaches its rated speed in \(5 \mathrm{s}\), and when the power is turned off, the unit coasts to rest in \(70 \mathrm{s}\). Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes \((a)\) in reaching its rated speed, \((b)\) in coasting to rest.

A painter is halfway up a 10 -m ladder when the bottom starts sliding out from under him. Knowing that point \(A\) has a velocity \(v_{A}=2 \mathrm{m} / \mathrm{s}\) directed to the left when \(\theta=60^{\circ}\), determine \((a)\) the angular velocity of the ladder, (b) the velocity of the painter.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.