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Statement-1 : Instantancous acceleration of a particle on a curved path may be along the instantancous velocity vector. Statement-2: Instantancous velocity is always tangential to the path.

Short Answer

Expert verified
Both statements are true; velocity is always tangential, and acceleration can align with velocity on curved paths.

Step by step solution

01

Evaluate Statement-1

Statement-1 claims that the instantaneous acceleration of a particle on a curved path may be along the instantaneous velocity vector. To evaluate this, understand that acceleration can have two components: tangential (along the velocity) and normal (perpendicular to velocity). For a curve, tangential acceleration occurs when the speed changes along the path. Therefore, it is possible that these components align when the path is not changing direction, hence acceleration may align with velocity.
02

Analyze Statement-2

Statement-2 asserts that instantaneous velocity is always tangential to the path of the particle. By definition, velocity is a vector that represents the rate of change of position. Since a curved path can be broken down into a series of tangents at any point, the velocity vector is inherently tangential to the path at each instant.
03

Conclusion from Steps

Both statements presented are correct. Statement-1 is possible under specific conditions where the change in velocity aligns with the movement of the curve. Statement-2 holds universally true since velocity, by definition, is tangential to the path.

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Key Concepts

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

Tangential acceleration
Tangential acceleration refers to the component of acceleration that acts along the direction of the velocity vector of a particle. This happens in scenarios where a particle, like a car on a track or a skier on a slope, speeds up or slows down. When the speed changes while moving along a path, tangential acceleration is at play. In a curved path, this tangential component can be tricky to visualize, as it impacts how the particle moves along the path itself. Imagine riding a bicycle and speeding up while going around a bend. Your change in speed along that curve is due to tangential acceleration, attempting to adjust the particle's speed without altering the direction. Understanding tangential acceleration is crucial because it helps describe how objects move through space and change their total velocity, not just direction. Note that while tangential acceleration can indeed be in the same direction as instantaneous velocity, it's not always the case. Sometimes, acceleration points in different directions, modifying both speed and direction simultaneously.
Instantaneous velocity
Instantaneous velocity is a core concept in understanding motion, particularly on a non-linear or curved path. It is defined as the velocity of a particle at a specific moment or instant. Unlike average velocity, which is calculated over a period, instantaneous velocity zooms in on a precise point in time. In essence, it is a vector that combines both speed and direction at a given instant. This concept is instrumental in physics because it provides the ability to describe motion precisely.
  • Speed Part: It tells you how fast a particle moves at that instant in time.
  • Direction Part: It indicates the pathway instantaneously traced by the particle.
For a particle navigating a curved path, like a roller coaster car, the idea of instantaneous velocity ensures you know how the car is moving at every twist and turn. And remember, this velocity vector is always tangent to the motion path, representing the direction the particle is moving at each point.
Curved path
A curved path refers to the trajectory that lacks a straight line, incorporating bends and twists. When a particle moves along such a path, its motion is subject to several forces and components. Unlike a straight path, a curve challenges prediction and requires careful examination to understand and analyze Key elements involved are:
  • Changing Direction: As the particle moves, its direction continuously changes, even if the speed remains constant.
  • Components of Motion: Alongside tangential acceleration, there is also normal acceleration (often directed towards the center of curvature), which causes the change in direction of the velocity vector.
For example, let's imagine a satellite orbiting a planet. Its path is curved due to gravitational forces, and its velocity direction continually adjusts to stay on the orbital path. This continuous adjustment manifests as acceleration components that must be understood to maintain accurate predictions of movement. Thus, grasping motion on a curved path is essential in fields like engineering or astronomy, where motion prediction and control are paramount.

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Most popular questions from this chapter

A plumb line is suspended from the roof of a railroad car and the car is moving on a circular road. The inclination of the string of the plumb line w.r. vertical direction is (a) \(\theta=\tan ^{-1}\left(v^{2} r g\right)\) (b) \(\theta=\tan ^{-1}\left(r g / v^{2}\right)\) (c) \(\theta=\tan ^{-1}\left(v^{2} / r g\right)\) (d) \(\theta=\operatorname{lan}^{-1}\left(v^{2} r / g\right)\)

A force produces an acecleration of \(4 \mathrm{~m} / \mathrm{scc}^{2}\) in body of mass \(m_{1}\) and the same forec produccs an acccleration of \(6 \mathrm{~m} / \mathrm{s}^{2}\) on a mass \(m_{2}\). If the forcc is applicd to a body of mass \(\left(m_{1}+m_{2}\right)\), its aceclcration will be \(a\). Then (a) \(\frac{m_{1}}{m_{2}}-\frac{3}{2}\) (b) \(\frac{m_{1}}{m_{2}}-\frac{2}{3}\) (c) \(a-2.4 \mathrm{~m} / \mathrm{scc}^{2}\) (d) \(\mathrm{a}-54 \mathrm{~m} / \mathrm{scc}^{2}\)

Two blocks of masses \(m\) and \(M\) are connccted by an incxicnsible light string. When a constant horizontal forec \(F\) acts on the block of mass \(M\), find the (a) tension in the string, (b) acceleration of the blocks, (c) notmal reactions on the blocks by the ground. \Lambdassuming \(m\) does not lose contact with the ground and ground is smooth.

Ten coins are placed on top of each other on a horizontal table. If the mass of each coin is \(10 \mathrm{gm}\) and acceleration due to gravity is \(10 \mathrm{~m} / \mathrm{s}^{2}\). What is the magnitude and direction of the force on the 7 th coin (counted from the bottom) due to all the coins above it? (a) \(0.7 \mathrm{~N}\) vertically downwards (b) \(0.7 \mathrm{~N}\) vertically upwards (c) \(0.3 \mathrm{~N}\) vertically downwards (d) \(0.3 \mathrm{~N}\) vertically upwards

\(\Lambda\) cat of mass \(m=1 \mathrm{~kg}\) climbs to a rope hung over a light frictionless pulley. The opposite end of the rope is tied to a weight of mass \(M=2 \mathrm{~m}\) lying on a smooth horizontal plane. What is the tension of the rope when the cat moves upwards with an acceleration \(a=2 \mathrm{~m} / \mathrm{s}^{3}\) relative to the rope'? Solution Let \(a\) be the absolute upward acceleration of the monkey and \(a\) ' be the absolute downward acceleration of the rope. \(a\) ' is also the tightward acceleration of \(M\). Then, \(b=a-\left(-a^{\prime}\right)\) (since relative acceleration is the vector difference between the absolute accelerations) or \(b-a=a^{\prime}\) Considering upward motion of the cat \(\quad T-m g=m a \ldots\) (i) Considering rightward motion of \(M\) \(T=M a^{\prime}=M(b-a) \quad \ldots(\) ii \()\) From (i) and (ii), we get \(T=\frac{m M}{m+M}(g+b)=\left(\frac{m \times 2 m}{m+2 m}\right)(10+2)=\frac{2 m}{3} \times 12=8 \mathrm{~N}\)

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