Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 557

A wheel rotates with a constant acceleration of $2.0 \mathrm{rad} / \mathrm{sec}^{2}$ If the wheel start from rest. The number of revolution it makes in the first ten seconds will be approximately. \(\\{\mathrm{A}\\} 8\) \\{B \\} 16 \(\\{\mathrm{C}\\} 24\) \(\\{\mathrm{D}\\} 32\)

Short Answer

Expert verified
The number of revolutions made by the wheel in the first 10 seconds is approximately \(\\{B\\} 16\).

Step by step solution

01

Write down relevant equations of motion for constant angular acceleration

There are two main equations that we can use for this problem: 1. Angular displacement equation (final angle, initial angle, initial angular velocity, time, and angular acceleration): \(\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\) 2. Angular velocity equation (final angular velocity, initial angular velocity, angular acceleration, and time): \(\omega = \omega_0 + \alpha t\) In this problem, we know that the initial angular velocity \(\omega_0 = 0\), the angular acceleration \(\alpha = 2.0 \, \text{rad/sec}^2\), and the time \(t = 10 \, \text{sec}\). We need to find the angular displacement in radians, \(\theta\), to calculate the number of revolutions.
02

Calculate the angular displacement using the first equation of motion

Using the equation \(\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2\), we can substitute our known values and calculate the angular displacement: \(\theta = 0 + 0 \cdot 10 + \frac{1}{2}(2.0) \cdot 10^2\) Calculating this value, we get: \(\theta = 100 \, \text{rad}\)
03

Convert angular displacement in radians to revolutions

Since there are 2π radians in one revolution, we can divide the angular displacement by 2π to get the number of revolutions: Number of revolutions = \(\frac{100 \, \text{rad}}{2π}\) Approximating π as 3.14, we have: Number of revolutions = \(\frac{100}{6.28} \approx 15.9\)
04

Choose the closest answer from the options

From our calculations, the number of revolutions is approximately 15.9. Looking at the answer choices, option B (16) is the closest to our calculated value. Therefore, the answer to this problem is: \(\\{B\\} 16\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Initial angular velocity of a circular disc of mass \(\mathrm{M}\) is \(\omega_{1}\) Then two spheres of mass \(\mathrm{m}\) are attached gently two diametrically opposite points on the edge of the disc what is the final angular velocity of the disc? \(\\{\mathrm{A}\\}[(\mathrm{M}+\mathrm{m}) / \mathrm{M}] \omega_{1}\) \(\\{\mathrm{B}\\}[(\mathrm{M}+4 \mathrm{~m}) / \mathrm{M}] \omega_{1}\) \(\\{\mathrm{C}\\}[\mathrm{M} /(\mathrm{M}+4 \mathrm{~m})] \omega_{1}\) \\{D\\} \([\mathrm{M} /(\mathrm{M}+2 \mathrm{~m})] \omega_{1}\)

The height of a solid cylinder is four times that of its radius. It is kept vertically at time \(t=0\) on a belt which is moving in the horizontal direction with a velocity \(\mathrm{v}=2.45 \mathrm{t}^{2}\) where \(\mathrm{v}\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) is in second. If the cylinder does not slip, it will topple over a time \(t=\) \(\\{\mathrm{A}\\} 1\) second \(\\{\mathrm{B}\\} 2\) second \\{C \(\\}\) \\} second \(\\{\mathrm{D}\\} 4\) second

A Pulley of radius \(2 \mathrm{~m}\) is rotated about its axis by a force \(F=\left(20 t-5 t^{2}\right) N\) where \(t\) is in sec applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is $10 \mathrm{KgM}^{2}$, the number of rotations made by the pulley before its direction of motion is reversed is : \(\\{\mathrm{A}\\}\) more than 3 but less then 6 \(\\{\mathrm{B}\\}\) more than 6 but less then 9 \(\\{\mathrm{C}\\}\) more than 9 \\{D \\} Less then 3

The M.I of a disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) about an axis passing through the centre \(\mathrm{O}\) and perpendicular to the plane of disc is \(\left(\mathrm{MR}^{2} / 2\right)\). If one quarter of the disc is removed the new moment of inertia of disc will be..... \(\\{\mathrm{A}\\}\left(\mathrm{MR}^{2} / 3\right)\) \(\\{B\\}\left(M R^{2} / 4\right)\) \(\\{\mathrm{C}\\}(3 / 8) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(3 / 2) \mathrm{MR}^{2}\)

A thin uniform rod \(A B\) of mass \(M\) and length \(L\) is hinged at one end \(\mathrm{A}\) to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end \(B\) strikes the floor is \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / \mathrm{L})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / \mathrm{L})}\) \(\\{C\\} \sqrt{(3 g / L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(\mathrm{g} / \mathrm{L})}\)
See all solutions

Recommended explanations on English Textbooks

Argumentative Essay

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Creative Story

Read Explanation

Lexis and Semantics

Read Explanation

Single Paragraph Essay

Read Explanation

Free Response Essay

Read Explanation
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.

Sign-up for free