Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 226

A particle is moving in a circle of radius \(\mathrm{R}\) with constant speed. It covers an angle \(\theta\) in some time interval. Find displacement in this interval of time. (A) \(2 \mathrm{R} \cos (\theta / 2)\) (B) \(2 \operatorname{Rsin}(\theta / 2)\) (C) \(2 \mathrm{R} \cos \theta\) (D) \(2 \mathrm{R} \sin \theta\)

Short Answer

Expert verified
The displacement of the particle during the time interval is \(\Delta AB = 2R\sin(\theta/2)\). Therefore, the correct answer is (B) \(2R \sin(\theta/2)\).

Step by step solution

01

Sketch the situation

Draw a circle with radius R, along with the initial position, final position, and the angle theta that the particle covers during the time interval. Label the initial position as A, the final position as B, and the center of the circle as O.
02

Find the displacement

In order to find the displacement (∆AB), we can use the law of cosines for triangles, which states that for a triangle with sides a, b, c and opposite angles A, B, C, we have: \(c^2 = a^2 + b^2 - 2ab \cdot \cos(C)\) In this problem, the triangle is OAB, where OA and OB are the radii of the circle, so OA = OB = R, and the angle AOB is given by theta. We want to find the length of the side AB, so we can apply the law of cosines: \(\Delta AB^2 = R^2 + R^2 - 2R \cdot R \cdot \cos(\theta)\)
03

Simplify the equation and find the value of displacement

Simplify the equation, we get: \(\Delta AB^2 = 2R^2 - 2R^2 \cdot \cos(\theta)\) Take the square root of both sides: \(\Delta AB = \sqrt{2R^2 - 2R^2 \cdot \cos(\theta)}\) Factor out the 2R^2: \(\Delta AB = R\sqrt{2(1 - \cos(\theta))}\) Using the trigonometric identity \(\cos(\theta) = 1 - 2\sin^2(\theta/2)\), we can rewrite the displacement as: \(\Delta AB = R\sqrt{2 \cdot 2\sin^2(\theta/2)}\) Further simplifying: \(\Delta AB = 2R\sin(\theta/2)\) So the displacement of the particle during the time interval is: \(\Delta AB = 2R\sin(\theta/2)\) Therefore, the correct answer is (B) \(2R \sin(\theta/2)\).

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

A freely falling stone crashes through a horizontal glass plate at time \(t\) and losses half of its velocity. After time \((t / 2)\) it falls on the ground. The glass plate is \(60 \mathrm{~m}\) high from the ground. Find the total distance travelled by the stone. \(\left[g=10 \mathrm{~ms}^{-2}\right]\) (A) \(120 \mathrm{~m}\) (B) \(80 \mathrm{~m}\) (C) \(100 \mathrm{~m}\) (D) \(140 \mathrm{~m}\)

$\quad\left(\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right) \cdot\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right)=\ldots$ (A) 0 (B) \(A^{2}+B^{2}\) (C) \(\sqrt{\left(A^{2}+B^{2}\right)}\) (D) \(\mathrm{A}^{2} \mathrm{~B}^{2}\)

Linear momentum of a particle is $(3 \mathrm{i} \wedge+2 \mathrm{j} \wedge-\mathrm{k} \wedge) \mathrm{kg} \mathrm{ms}^{-1}$. Find its magnitude. (A) \(\sqrt{14}\) (B) \(\sqrt{12}\) (C) \(\sqrt{15}\) (D) \(\sqrt{11}\)

Find a unit vector from the followings. (A) \(\hat{\imath}+\hat{j}\) (B) \(\hat{\imath}-\hat{j}\) (C) \((1 / \sqrt{2}) \hat{\imath}+(1 / \sqrt{2}) \hat{\jmath}\) (D) \((1 / \sqrt{2}) \hat{1}-(1 / 2) \hat{\jmath}\)

Find a unit vector perpendicular to both \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) $\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) / \mathrm{AB}\right]$ (B) $\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{-}\right) /(\mathrm{AB} \sin \theta)\right]$ (C) $\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \cos \theta)\right]$ (D) $\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \sin \theta)\right]$
See all solutions

Recommended explanations on English Textbooks

Phonetics

Read Explanation

Essay Prompts

Read Explanation

Research and Composition

Read Explanation

Creative Story

Read Explanation

Sociolinguistics

Read Explanation

Lexis and Semantics Summary

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