/*! 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 40 Two particles start moving on ci... [FREE SOLUTION] | 91Ó°ÊÓ

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Two particles start moving on circles of radius \(R_{1}=1 \mathrm{~m}\) and \(R_{2}=3 \mathrm{~m}\) shown with velocity \(v_{1}=2 \mathrm{~m} / \mathrm{s}, v_{2}=15 \mathrm{~m} / \mathrm{s}\) respectively. The minimum time after which the particles will again be collinear with centre is equal to (1) \(\frac{2 \pi}{3} \mathrm{sec}\) (2) \(\frac{2 \pi}{5} \mathrm{sec}\) (3) \(2 \pi \mathrm{sec}\) (4) \(\frac{\pi}{3} \mathrm{sec}\)

Short Answer

Expert verified
The minimum time is \(\frac{2\pi}{3}\, \mathrm{sec}\).

Step by step solution

01

Determine the angular velocities

The angular velocity \(\omega\) for a particle moving in a circular path can be calculated using the formula \(\omega = \frac{v}{R}\), where \(v\) is the linear velocity and \(R\) is the radius of the circle. For particle 1, \(\omega_1 = \frac{v_1}{R_1} = \frac{2}{1} = 2\, \text{rad/s}\). For particle 2, \(\omega_2 = \frac{v_2}{R_2} = \frac{15}{3} = 5\, \text{rad/s}\).
02

Find the angular velocities' difference

The difference in angular velocities, \(\Delta \omega\), determines how quickly the particles will realign. It's given by \(\Delta \omega = |\omega_1 - \omega_2| = |2 - 5| = 3\, \text{rad/s}\).
03

Determine the period of realignment

In circular motion, particles will be collinear with the center again after their angular positions differ by a multiple of \(2\pi\). The minimum time \(T\) for this can be found using \(\Delta \omega \cdot T = 2\pi\). Solving for \(T\) gives \(T = \frac{2\pi}{\Delta \omega} = \frac{2\pi}{3}\, \mathrm{s}\).

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Key Concepts

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

Angular Velocity
Angular velocity is a crucial concept when discussing circular motion. It measures how fast an object rotates or revolves relative to another point. Think of it as the speed of a spin!
Angular velocity is denoted by the Greek letter \(\omega\) and is usually expressed in radians per second (rad/s). To compute the angular velocity for a particle moving along a circle, we use the formula:
  • \(\omega = \frac{v}{R}\), where \(v\) is the linear velocity and \(R\) is the radius of the circle.
For instance, in the problem, the angular velocity of particle 1 is calculated as \(\omega_1 = \frac{2}{1} = 2\, \text{rad/s}\), which indicates how quickly the particle makes a complete rotation around the circle. Similarly, the angular velocity of particle 2 is \(\omega_2 = \frac{15}{3} = 5\, \text{rad/s}\).
Remember, angular velocity is vital in determining how particles move in synchrony when they're on different circular paths.
Linear Velocity
While angular velocity refers to rotational speed, linear velocity describes how fast an object moves in a straight line, especially along the edge of a circle. In circular motion, the direction of linear velocity is always tangent to the circle.
For an object moving in a circle, linear velocity is related to angular velocity and the radius of the circle:
  • \(v = \omega \cdot R\), where \(\omega\) is the angular velocity, and \(R\) is the radius of the circle.
In our exercise, particle 1 has a linear velocity of \(2\, \text{m/s}\), and particle 2 has \(15\, \text{m/s}\).
The calculation of linear velocity helps us understand how quickly an object might travel if it continues in its circular path for a certain time.
Period of Realignment
The concept of the period of realignment is intriguing in circular motion. It tells us the time it takes for two or more revolving particles to line up together based on their paths and speeds.
In this context, two particles will realign their positions with respect to a central point when their angular positions differ by a multiple of \(2\pi\) radians.
The time it takes for this to occur, \(T\), is given by:
  • \(T = \frac{2\pi}{\Delta \omega}\), where \(\Delta \omega\) is the difference in angular velocities of the particles.
From our task, \(\Delta \omega = |2 - 5| = 3\, \text{rad/s}\). Therefore, the minimum realignment period is \(T = \frac{2\pi}{3}\, \text{s}\), meaning that every \(\frac{2\pi}{3}\) seconds, the particles will be in straight line with the center again.
Radius of Circle
In circular motion, the radius of a circle is a fundamental element that affects both linear and angular velocities. It is the distance from the center of the circle to any point on its circumference.
The radius plays a significant role in determining how an object moves because:
  • Angular velocity \(\omega\) is influenced by the radius through the formula \(\omega = \frac{v}{R}\).
  • Linear velocity \(v\) is proportional to both angular velocity and radius \(v = \omega \cdot R\).
In this exercise, particle 1 circles a radius of \(1\, \text{m}\), while particle 2 has a larger circle with a radius of \(3\, \text{m}\).
The different radii impact how each particle moves relative to time, affecting their speed and how they maintain synchronicity or alignment with each other.

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Most popular questions from this chapter

Two strings of length \(l=0.5 \mathrm{~m}\) each are connected to a block of mass \(m=2 \mathrm{~kg}\) at one end and their ends are attached to the point \(A\) and \(B, 0.5 \mathrm{~m}\) apart on a vertical pole which rotates with a constant angular velocity \(\omega=7 \mathrm{rad} / \mathrm{s}\). Find the ratio \(T_{1} / T_{2}\) of tension in the upper string \(\left(T_{1}\right)\) and the lower string \(\left(T_{2}\right)\). [Use \(\left.g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right]\)

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A particle is moving with a velocity of \(\vec{v}=(3 i+4 t j) \mathrm{m} / \mathrm{s}\) Find the ratio of tangential acceleration to that of normal acceleration at \(t=1 \mathrm{sec}\) (1) \(4 / 3\) (2) \(3 / 4\) (3) \(5 / 3\) (4) \(3 / 5\)

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