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Chapter 3: Problem 38

\(\Lambda\) particle moves in a circular path of radius \(R\) with an angular velocity \(\omega \mathrm{rad} / \mathrm{s}\) given by \(\omega=a-b i\) where \(a\) and \(b\) are positive constants and \(t\) is time. Find the magnitude of the acceleration of the particle after time \(2 a / b\).

Short Answer

Expert verified
The magnitude of the acceleration is \((a^2 + b^2)R\).

Step by step solution

01

Understand angular velocity

The angular velocity, represented as \(\omega\), is given by \(a - bi\). In this complex form, \(a\) is the real part and \(-bi\) is the imaginary part. This represents motion with constant angular speed \( |a| \) but the presence of an imaginary part indicates a change in direction.
02

Calculate the real angular velocity

Since the angular component is not purely real due to the presence of \(-bi\), we consider the real part for practical angular motion, \(\omega = a\).
03

Find the angular position after time 2a/b

The angular position \( \theta \) at any time \( t \) is given by \( \theta = \omega \cdot t = (a-bi)t \). At time \( t = \dfrac{2a}{b} \), \( \theta = a \cdot \dfrac{2a}{b} - b i \cdot \dfrac{2a}{b} = \dfrac{2a^2}{b} - 2ai \).
04

Determine the magnitude of velocity

Velocity \( v \) is related to angular velocity by \( v = \omega \times R \). However, since \( \omega \) has an imaginary component, only the real part impacts physical velocity. Thus, \( v = aR \) and imaginary parts don't directly affect linear speed.
05

Calculate the total acceleration

For circular motion, acceleration has two components: centripetal \( a_c = \omega^2 R \) and tangential \( a_t = \dfrac{dv}{dt} \). Since angular speed is constant at instant from real component, and \( \omega = a \), the tangential component is 0. The magnitude of centripetal acceleration is \( |(a-bi)^2|R = (a^2 + b^2)R \). So, the total acceleration is \((a^2 + b^2)R\).

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Key Concepts

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

Understanding Angular Velocity
Angular velocity is a measure of how fast an object rotates around a circle. It describes the speed of rotation and its direction. In circular motion problems, it's usually symbolized as \( \omega \). When expressed in a standard form, angular velocity is a real number. However, it can also be described using complex numbers, as shown by \( \omega = a - bi \), where \( a \) and \( b \) are constants. Here, \( a \) is the real part indicating actual rotational speed, and \( -bi \) is the imaginary part affecting the direction of rotation.
  • Real Part (\( a \)): Indicates the basic constant speed of the rotation.
  • Imaginary Part (\( -bi \)): Suggests a shift in direction, adding complexity to the motion.
Understanding angular velocity helps in determining how fast an angle is covered in circular motion.
Exploring Centripetal Acceleration
Centripetal acceleration keeps an object moving in a circular path. It points towards the center of the circle, ensuring the motion doesn鈥檛 spiral outward. For any rotating object, centripetal acceleration can be computed by the formula:\[ a_c = \omega^2 R \]Here, \( \omega \) is the angular velocity and \( R \) is the radius of the circle. When angular velocity includes complex elements, ensure to calculate the squared magnitude:\[ \text{Magnitude of } (a - bi)^2 = a^2 + b^2 \]
  • \( a^2 \): Represents the contribution of the real speed part to acceleration.
  • \( b^2 \): Accounts for directional changes impacting acceleration.
Therefore, centripetal acceleration here adapts to the complexity introduced by the imaginary element, ensuring the path remains circular.
Complex Numbers in Motion
Complex numbers often appear in physics to depict multi-dimensional issues effortlessly. In the context of circular motion, they convey shifts in direction or varying speed factors, expressed as combinations of real and imaginary numbers such as \( a - bi \).
  • Real Component (\( a \)): Contributes to the actual speed in the circular path.
  • Imaginary Component (\( -bi \)): Adds depth to motion, indicating direction changes.
Finite time considerations using complex numbers aid in defining complete circular paths by merging linear speed and directional shifts. They play a crucial role in moving from linear insights to rotationally complex analyses.
Angular Position with Time
Angular position refers to the angle an object has rotated about a circle at any time. It describes the object's orientation over time, calculated by multiplying angular velocity \( \omega \) and time \( t \):\[ \theta = \omega \cdot t \]When \( \omega \) is given as \( a - bi \) and time is \( t = \frac{2a}{b} \), the angular position is:\[ \theta = a \cdot \frac{2a}{b} - b i \cdot \frac{2a}{b} = \frac{2a^2}{b} - 2ai \]
  • \( \frac{2a^2}{b} \): Reflects the linear distance moved along the circle based on constant speed.
  • \( -2ai \): Denotes directional adjustments over the given time.
Angular position, calculated using both real and imaginary parts of angular velocity, showcases the particle鈥檚 precise location and direction at any time.

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Most popular questions from this chapter

Statement-1 : In projectile motion if angle of projection is made \(60^{\circ}\) instead of \(30^{\circ}\), the minimum kinctie cnergy will become \(\frac{1}{3}\) limes. Initial speed remaining the same. Statement-2 : In projectile motion, minimum kinetic energy is proportional to \(\cos ^{2} \theta\), where \(\theta\) is angle of projection.

\(\Lambda\) bus is going in south with a speed of \(5 \mathrm{~m} / \mathrm{s}\). To a man sitting in bus a car appears to move towards west with a speed of \(2 \sqrt{6} \mathrm{~m} / \mathrm{s}\). What is the actual speed of the car? (a) \(4 \mathrm{~ms}^{-1}\) (b) \(3 \mathrm{~ms}^{-1}\) (c) \(7 \mathrm{~ms}^{-1}\) (d) None

At \(t=2\) sec, ta projectile is moving upwards. Its sped is \(20 \mathrm{~m} / \mathrm{s}\) and it is making an angle \(30^{\circ}\) with horizontal. What is angle o[ projoction of the projectile from gound? \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(\operatorname{lan}^{-1}(2)\) (b) \(45^{\circ}\) (c) \(60^{\circ}\) (d) \(\tan ^{-1}(3)\)

Statement-1 : Displacement of a particle is always less than or cqual to the distance travel by the particle. Statement- \(2:\) For a particle moving with constant velocity the distance and displacement has equal magnitude.

\(\Lambda\) particle moves along the trajectory of parabola \(y=a x^{2} .\) \Lambdassuming speed to be constant, find the radius of curvature at the origin. (a) \(\frac{1}{2 a}\) (b) \(\frac{1}{a}\) (c) \(\frac{1}{2 a x}\) (d) \(\frac{2}{a}\)
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