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Chapter 3: Problem 12

Statement-1: In a non-uniform circular motion tangential acecleration arises due to change in magnitude of velocity. Statement-2 : In a non-uniform circular motion centripctal acecleation produce duc to change in direction of velocity.

Short Answer

Expert verified
Both statements are true.

Step by step solution

01

Understanding Non-uniform Circular Motion

In a non-uniform circular motion, the speed of the object is not constant, meaning it changes as the object moves around the circle. This change in speed indicates that there is an acceleration component along the direction of the velocity (tangential acceleration).
02

Tangential Acceleration

Tangential acceleration arises when there is a change in the magnitude of velocity. In non-uniform circular motion, either the object speeds up or slows down, resulting in a change in velocity magnitude.
03

Direction Change and Centripetal Acceleration

Even if the speed remains constant, an object moving in a circle is accelerating due to the continuous change in direction. This requires centripetal acceleration, which acts towards the center of the circle, maintaining the circular path.
04

Evaluating Statement 1

In Statement 1, it is correctly stated that tangential acceleration arises from a change in the magnitude of velocity. Therefore, this statement is true.
05

Evaluating Statement 2

Statement 2 correctly mentions that centripetal acceleration is due to the change in direction of velocity, which is characteristic of circular motion. Thus, this statement is also true.
06

Conclusion

Both statements discuss different components of acceleration in non-uniform circular motion and are both true based on their respective explanations of tangential and centripetal accelerations.

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Key Concepts

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

Tangential Acceleration
Tangential acceleration is an essential concept in non-uniform circular motion. It occurs when there is a change in the magnitude of velocity. Simply put, it is the component of acceleration that causes an object to speed up or slow down along a circular path. This type of acceleration acts in the direction of the velocity vector.

Imagine a car moving on a circular track. As the car accelerates, its speed increases; when it decelerates, its speed decreases. This change in speed represents tangential acceleration.
  • The formula for tangential acceleration is given by:
    \[ a_t = \frac{dv}{dt} \]
    where \( a_t \) is the tangential acceleration, and \( \frac{dv}{dt} \) is the rate of change of velocity.
  • Tangential acceleration affects the speed but not the direction of the object.
Centripetal Acceleration
Centripetal acceleration is crucial for understanding circular motion, even when moving at a constant speed. It arises from the change in direction of the velocity vector, not its magnitude. In circular motion, this acceleration points towards the center of the circle.

Let's visualize a ball tied to a string and swung in a circle. Despite a constant speed, the ball continually changes direction, necessitating centripetal acceleration. This keeps the ball on its circular path.
  • The formula for centripetal acceleration is:
    \[ a_c = \frac{v^2}{r} \]
    where \( a_c \) is the centripetal acceleration, \( v \) is the velocity magnitude, and \( r \) is the radius of the circular path.
  • Centripetal acceleration does not alter the object's speed, only the direction.
Velocity Change
Velocity change in circular motion is a comprehensive concept. It encompasses both changes in speed and direction. When an object moves in a non-uniform circular path, its velocity is continuously changing. This change gives rise to both tangential and centripetal accelerations.

  • The concept of velocity in circular motion is twofold:
    • Magnitude change: Corresponds to tangential acceleration, affecting how fast or slow an object moves along its path.
    • Direction change: Corresponds to centripetal acceleration, altering the trajectory but not affecting speed.
  • Understanding velocity change is vital for grasping how objects behave in circular motion.
Mastering tangential and centripetal accelerations helps in predicting and solving problems related to circular dynamics, making velocity change a critical concept for students.

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Most popular questions from this chapter

\(\Lambda\) motorboat covers a given distance in \(6 \mathrm{~h}\) moving downstream on a river. It covers the same distance in \(10 \mathrm{~h}\) moving upstream. Find the time it takes to cover the same distance in still water.

A particle moves along positive branch of the curve \(y=x / 2\); where \(x=t^{3} / 3, x\) and \(y\) are measured in metres and \(t\) in seconds, then (a) the velocity of particle at \(t=1 \mathrm{~s}\) is \(\hat{i}+(1 / 2) \hat{j}\) (b) the velocity of particle at \(t=1 \mathrm{~s}\) is \((1 / 2) \hat{i} \mid \hat{j}\) (c) the acecleration of particle at \(t=2 \mathrm{~s}\) is \(2 \hat{i}+\hat{j}\) (d) the acceleration of particle at \(t=2 \mathrm{~s}\) is \(\hat{i}\) ? \(2 \hat{j}\)

A particle is projected vertically upward with velocity \(40 \mathrm{~m} / \mathrm{s}\). Tind the displaccment and distancc travellod by the particle in (a) \(2 \mathrm{~s},(\mathrm{~b}) 4 \mathrm{~s}\) and \((\mathrm{c})\) os \(\left(\right.\) Take \(\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\).

A particle is to be projected so as to just pass through three equal rings of diameter \(d\) and placed in parallel vertical planes at distances \(a\) apart with their highest points at a height \(h\) above the point of projection. Prove that the clcvation of projection is tan \(\frac{\sqrt{h d}}{a}\).

\Lambda particle moving in \(x-y\) plane is at origin at time \(t=0 .\) Velocity of the particle is \((2 \vec{i}-4 \hat{j}) \mathrm{m} / \mathrm{s}\) and acceleration \(i s(4 \hat{i} \mid \hat{j}) \mathrm{m} / \mathrm{s}^{2}\). Find at \(t=2 \mathrm{~s}\) (a) velocity of particle and (b) coordinates of particle.
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