Derivation of Centripetal Acceleration - Detailed Guide

Derivation of Centripetal Acceleration - Detailed Guide

Edited By Vishal kumar | Updated on Jul 02, 2025 05:04 PM IST

Imagine driving around a curve and feeling a pull outward—that's centripetal acceleration in action, a key concept in circular motion. It helps explain how Forces work on a vehicle during turns. Understanding this can make driving safer, especially when navigating curves at higher speeds. In this report, we explore the role of centripetal acceleration, factors that affect it like speed and friction, and practical tips for handling turns. In this article we will be studying centripetal acceleration is, centripetal acceleration formula, centripetal acceleration derivation, centripetal force, define centripetal acceleration, direction of centripetal acceleration and centrifugal acceleration formula. Define Centripetal Acceleration.

This Story also Contains
  1. Define Centripetal Acceleration and Derive an Expression
  2. Centripetal Acceleration Formula:
  3. Centripetal Acceleration Derivation
  4. Centripetal Force
  5. Centripetal Force Derivation
  6. Centrifugal Force Formula Derivation
  7. Centripetal Force vs centrifugal Force
  8. Problem-Solving Strategy
  9. Weitage of Centripetal Acceleraion in Diffrent Exam
Derivation of Centripetal Acceleration - Detailed  Guide
Derivation of Centripetal Acceleration - Detailed Guide

This concept, covered in Class 11 Physics under circular motion, derivation of centripetal acceleration is essential for board exams and competitive exams like JEE Main, NEET, BITSAT, and others. Between 2013 and 2023, five questions on this topic have appeared in JEE Main.

Define Centripetal Acceleration and Derive an Expression

Centripetal acceleration is the acceleration of a body that is travelling across a circular path. When a body undergoes a circular motion, its direction constantly changes and thus its velocity changes (velocity is a vector quantity) which produces an acceleration. The centripetal acceleration ac is given by the square of speed v divided by the distance "r".
Centripetal acceleration is the acceleration of a body that is travelling across a circular path.  When a body undergoes a circular motion, its direction constantly changes and thus its velocity changes (velocity is a vector quantity) which produces an acceleration. The acceleration and hence force is towards the centre of the circle. The magnitude of centripetal acceleration ac is given by the square of speed v divided by the distance r;1726225427674

Here, a question arises that why is it called centripetal acceleration while the speed of rotation remains constant?

The question is valid. Now we’re providing you with the answer to the query, and it is because speed is a scalar quantity. At the same time, in rotational motion, there is no actual speed of rotation because it includes direction as well; instead, there is rotational velocity(vector quantity). Hence, velocity variation leads to the term named centripetal acceleration.

Centripetal Acceleration Formula:

$$
a_c=\frac{v^2}{r}
$$

where:
-.$a_c$ is the centripetal acceleration,
- $v$ is the velocity of the object,
- $r$ is the radius of the circular path.

This is the required Centripetal acceleration Formula.

  • Centripetal acceleration unit: metre per second squared (m/s2).
  • Centripetal acceleration Dimensional formula:

The dimensional formula of centripetal acceleration is given as:

Dimensional formula of radius = M0L1T0

Dimensional formula of Velocity = M0L1T-1

Substituting the above in the equation of the centripetal formula a= v2/r, we get:

Dimensional formula of Centripetal acceleration = M0L1T-2

Note: The force causing this acceleration is also directed towards the centre of the circle and is named centripetal force.

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Centripetal Acceleration Derivation

Consider a body of mass ‘m’ moving on the circumference of a circle of radius ‘r’ with a velocity ‘v’. A force F is then applied to the body. And this force is given by

F = ma

Centripetal Acceleration Derivation1726225427712

Where, a= acceleration which is given by the rate of change of velocity with respect to time.

Consider the triangles $\triangle O A B$ and $\triangle P Q R$, then:

$$
\Delta v_{A B}=\frac{v}{r}
$$
Clearly, from the diagram (not shown here), the arc length $A B=v \Delta t$.
Now, we have:

$$
\frac{\Delta v}{v \Delta t}=\frac{v}{r}
$$
Simplifying this:

$$
\frac{\Delta v}{\Delta t}=\frac{v^2}{r}
$$
Thus, the centripetal acceleration equation is:

$$
a=\frac{v^2}{r}
$$

Note: The direction of centripetal acceleration(& force) is towards the centre of the circle.

Now, let's move to another part of this article which is Centripetal Force

What is the direction of centripetal acceleration?

The direction of centripetal acceleration and also force is towards the centre of the circle.

Centripetal Force

It is the force that acts on a body undergoing circular motion and is directed towards the centre of the rotation(or the circle).

Centripetal Force Digrame1726225427749

Centripetal Force Derivation

Centripetal force is the net force causing uniform circular motion.

According to Newton’s laws of motion,

The force is given by $F=m a$, where:
- $m$ is the mass of the body,
- $a$ is the acceleration of the body.

In the case of uniform circular motion, the acceleration is the centripetal acceleration, given by:

$$
a_c=\frac{v^2}{r}
$$

where:
- $v$ is the linear velocity of the object,
- $r$ is the radius of the circular path.

Now, using angular velocity $(\omega)$, centripetal acceleration can also be expressed as:

$$
a_c=r \omega^2
$$

where:
- $\omega$ is the angular velocity of the object,
- $r$ is the radius of the circular path.

Thus, the centripetal force formula using linear velocity is:

$$
F_c=\frac{m v^2}{r}
$$

Here:
- $F_c$ is the centripetal force,
- $m$ is the mass of the object,
- $v$ is the linear velocity,
- $r$ is the radius of the circular path.

The centripetal force formula in terms of angular velocity is given by:

$$
F_c=m r \omega^2
$$

where:
- $F_c$ is the centripetal force,
- $m$ is the mass of the object,
- $r$ is the radius of the circular path,
- $\omega$ is the angular velocity.

Thus, the expression for centripetal force in terms of angular velocity is:

$$
F_c=m r \omega^2
$$

Or,

We can say that $\frac{m v^2}{r}$is the formula of centripetal acceleration.

It is also known as angular centripetal force/derivation of centrifugal force class 11

Centrifugal Force Formula Derivation

After centripetal force derivation, now in this section, we are going to study centrifugal force derivation and its formula which is one of the important concepts for board exams.

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Centrifugal Force

$$
F=m \times \frac{v^2}{r}
$$

2. In terms of angular velocity $\omega$ :

$$
F=m \times \frac{(r \omega)^2}{r}=m \times r \omega^2
$$

Thus, the formula for centripetal force can be expressed as either:

$$
F=m \times \frac{v^2}{r}
$$

or

$$
F=m \times r \omega^2
$$

where $m$ is the mass, $v$ is the linear velocity, $r$ is the radius of the circular path, and $\omega$ is the angular velocity.

NOTE:

  • For a particle in circular motion, the centripetal acceleration is:
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$$
a_c=\frac{v^2}{r} \quad \text { or } \quad a_c=r \omega^2
$$

  • The expression $m \omega^2 r$ represents the centripetal force.
  • The formula for centrifugal force (which is equal in magnitude but opposite in direction to centripetal force) is:

$$
F=m \omega^2 r=\frac{m v^2}{r}
$$

Derive v= r ω

Derive v= r ω1726225427825

Let us consider a body rotating about an axis that is passing through O point and also perpendicular to the plane.

Let us suppose, P be the position of a particle inside the body. If the body rotates an angle 0 in a time ‘t’, the particle at which is at P reaches P′

The displacement along the circular path is:

$$
P P^{\prime}=r \theta
$$

The velocity is:

$$
v=\frac{P P^{\prime}}{t}=\frac{r \theta}{t}
$$
Since $\frac{\theta}{t}=\omega$ (angular velocity), we have:

$$
v=r \omega
$$

Using the relationship $v=r \omega$, we can write:

$$
v^2=(r \omega)^2=r^2 \omega^2
$$

Substitute $v^2$ into the acceleration formula:

$$
a=\frac{r^2 \omega^2}{r}
$$

Simplifying:

$$
a=r \omega^2
$$
\begin{aligned}
&\text { Thus, the centripetal acceleration can be expressed as: }\\
&a=\frac{v^2}{r}=r \omega^2
\end{aligned}

Also read







Centripetal Force vs centrifugal Force

In the previous section, we derive an expression for centripetal acceleration. A key concept that often comes to mind is the difference between centripetal and centrifugal forces. Below, we outline some of the fundamental differences between these two forces. Please review them carefully.

Centripetal Force:

Definition: Centripetal force is the force that acts on an object moving in a circular path, directed towards the centre of the circle. It keeps the object moving in a curved trajectory rather than in a straight line.

Nature: A real force that acts on the object.

Direction: Always directed towards the centre of the circular path.

Formulae: $F_c=\frac{m v^2}{r}$

Examples:

1. The tension in a string when swinging a ball in a circular motion.

2. The gravitational force acting on planets keeps them in orbit.

Centrifugal Force:

Definition: Centrifugal force is a pseudo or fictitious force that appears to act on an object when it is observed from a rotating reference frame. It seems to push the object away from the centre of rotation.

Nature: Not a real force; it is an apparent force observed in a non-inertial (rotating) frame of reference.

Direction: Directed away from the centre of the circular path, opposite to the direction of centripetal force.

Formula: Fcentrifugal =$-\frac{m v^2}{r}=-m r \omega^2$

Examples:

1. The sensation of being "thrown" outward when a car makes a sharp turn.

2. The apparent force felt by passengers in a spinning amusement park ride.

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Problem-Solving Strategy

1. Identify the plane of circular motion.

2. Locate the centre of rotation and calculate the radius.

3. Make F.B.D.

4. Resolve force along the radial direction and along the direction perpendicular to it.

5. The net force along the radial direction is mass times the radial acceleration is no separate force like Tension, Weight, Spring force, Normal reaction, Friction etc. In fact any of these or their combination may play a role of centripetal force.

After going through the derivation of centripetal acceleration now you will able to solve questions based on this concept and this derivation is also very important for board exams.

Weitage of Centripetal Acceleraion in Diffrent Exam

ExamWeight of Centripetal AccelerationRemarks
JEE Main and JEE AdvancedModerate to High

Often tested in mechanics, rotational dynamics, and central forces problems.

NEET (Medical Entrance Exam)Moderate

Related to biomechanics and motion in circular paths, but less focus compared to engineering exams.

CUET (Common University Entrance Test)Low to Moderate

Tested in science-oriented courses; typically involves straightforward, application-based questions.

GATE (Graduate Aptitude Test in Engineering)High

Critical in dynamics and machine theory, particularly in mechanical and civil engineering sections.

Class 12 Board Exams (CBSE/State Boards)Moderate

Part of the physics syllabus in mechanics and circular motion chapters; direct formula-based questions.

Olympiads (National and International)High

Deeply explored in challenging mechanics problems requiring advanced understanding and applications.


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Frequently Asked Questions (FAQs)

1. Centripetal acceleration formula derivation.

We start with the formula for centripetal force:

$$
F=\frac{m v^2}{r}
$$
Also, from Newton's 2nd law:

$$
F=m a
$$
We can express acceleration as:

$$
a=\frac{F}{m}
$$
Substituting the value of $F$ from the centripetal force equation:

$$
a=\frac{\frac{m v^2}{r}}{m}
$$
Simplifying it:

$$
a=\frac{v^2}{r}
$$
Thus, the centripetal acceleration is:

$$
a=\frac{v^2}{r}
$$

2. Centripetal acceleration formula proof.

 Proof:

1. $F=\frac{m v^2}{r}$ (using centripetal force)

2. $F=m a$ (using Newton's 2nd law)

We can rewrite this as:

3. $a=\frac{F}{m}$

Substituting the value of $F$ from equation (1) into equation (3):

4. $a=\frac{m v^2}{r} \div m$

Simplifying:

5. $a=\frac{v^2}{r}$

This is the required expression for acceleration in circular motion.

3. Define centripetal acceleration derive an expression for it./What is centripetal acceleration derive an expression for it.

Centripetal acceleration is the acceleration of a body that is travelling across a circular path. When a body undergoes a circular motion, its direction constantly changes and thus its velocity changes (velocity is a vector quantity) which produces an acceleration.Derivation:

$F=\frac{m v^2}{r}$ (using centripetal force)

$F=m a$ (using Newton's second law)

We can rewrite this as:

$a=\frac{F}{m}$

Substituting the value of $F$ 

\$a=\frac{m v^2}{r} \div m$

$a=\frac{v^2}{r}$

Thus, the centripetal acceleration is given by $a=\frac{v^2}{r}$.

4. Derive an expression for centripetal force class 11. / Derivation of centripetal force class 11.

Derivation for centripetal force is given in this article.

5. Why doesn't an object experiencing only centripetal acceleration ever reach the center of its circular path?
An object with only centripetal acceleration maintains its circular path because the acceleration changes only the direction of velocity, not its magnitude. To reach the center, the object would need a component of acceleration opposite to its velocity to reduce its speed.
6. Why is understanding centripetal acceleration important for designing roller coasters?
Understanding centripetal acceleration is crucial for roller coaster design to ensure passenger safety and comfort. It helps engineers calculate the forces experienced in loops and turns, determine safe speeds, and design appropriate restraint systems.
7. Why is it important to consider centripetal acceleration in the design of space stations?
In space stations, artificial gravity can be created by rotation. Understanding centripetal acceleration is crucial for determining the rotation rate and radius needed to simulate Earth-like gravity, ensuring comfort and safety for astronauts.
8. How does centripetal acceleration affect the apparent weight of a person in a centrifuge?
In a centrifuge, centripetal acceleration increases the apparent weight of a person. The total force experienced is the vector sum of their weight and the centripetal force, resulting in a feeling of increased pressure towards the outer wall of the centrifuge.
9. What is the relationship between centripetal acceleration and angular acceleration?
While centripetal acceleration changes the direction of velocity in circular motion, angular acceleration changes the magnitude of angular velocity. They are distinct concepts, but both can be present in non-uniform circular motion.
10. What is the formula for centripetal acceleration?
The formula for centripetal acceleration is a = v²/r or a = ω²r, where 'a' is centripetal acceleration, 'v' is tangential velocity, 'r' is the radius of the circular path, and 'ω' is angular velocity.
11. How is centripetal acceleration related to angular velocity?
Centripetal acceleration is directly proportional to the square of angular velocity (ω). The relationship is expressed as a = ω²r, where 'r' is the radius of the circular path.
12. What is the difference between average and instantaneous centripetal acceleration?
In uniform circular motion, the instantaneous and average centripetal accelerations are the same, always pointing towards the center with constant magnitude. In non-uniform circular motion, the instantaneous acceleration can vary, while the average is taken over a period of time.
13. Why is centripetal acceleration always directed towards the center of the circle?
Centripetal acceleration is always directed towards the center because this is the direction in which the velocity vector must change to maintain circular motion. Any other direction would result in a deviation from the circular path.
14. How does the derivation of centripetal acceleration use vector calculus?
The derivation uses vector calculus to analyze the change in the velocity vector over time. It involves taking the derivative of the position vector with respect to time twice, which leads to the centripetal acceleration vector pointing towards the center of the circle.
15. Why does centripetal acceleration increase as the radius of the circular path decreases?
As the radius decreases, the object must change its direction more rapidly to maintain circular motion. This requires a greater centripetal acceleration, as reflected in the inverse relationship between acceleration and radius in the formula a = v²/r.
16. How is centripetal acceleration derived from the definition of acceleration?
Centripetal acceleration is derived by considering the change in velocity vector over a small time interval in circular motion. The derivation involves vector calculus, showing that the acceleration points towards the center and has a magnitude of v²/r.
17. Why is centripetal acceleration constant in uniform circular motion?
In uniform circular motion, the speed (magnitude of velocity) remains constant. Since centripetal acceleration depends on velocity squared and radius (a = v²/r), and both these quantities are constant, the centripetal acceleration is also constant.
18. How does the period of rotation relate to centripetal acceleration?
The period of rotation (T) is inversely related to centripetal acceleration. As the period decreases (faster rotation), the centripetal acceleration increases. This relationship can be expressed as a = 4π²r/T².
19. How does centripetal acceleration affect the total acceleration of an object in circular motion?
The total acceleration of an object in circular motion is the vector sum of centripetal acceleration and any tangential acceleration. In uniform circular motion, the total acceleration is just the centripetal acceleration, as there is no tangential component.
20. How does the concept of centripetal acceleration apply to the motion of water in a curved hose?
Water moving through a curved hose experiences centripetal acceleration. The walls of the hose provide the necessary force to change the water's direction, resulting in pressure on the outer wall of the curve.
21. Why is it incorrect to think of centripetal acceleration as "pulling" an object towards the center?
Centripetal acceleration doesn't "pull" an object; it describes the object's motion. The acceleration is a result of a force (like tension or gravity) that changes the object's velocity direction. Thinking of it as a "pull" can lead to misconceptions about the forces involved.
22. How does the concept of centripetal acceleration apply to the motion of electrons in a cyclotron?
In a cyclotron, charged particles like electrons move in a spiral path due to a magnetic field. The magnetic force provides the centripetal acceleration necessary for the circular component of this motion, allowing the particles to be accelerated to high energies.
23. Why is it important to consider centripetal acceleration when analyzing the motion of fluids in curved pipes?
Understanding centripetal acceleration is crucial for analyzing fluid motion in curved pipes. It helps explain pressure differences between the inner and outer walls of the curve and is important for designing efficient piping systems in various industries.
24. What is the significance of centripetal acceleration in understanding the stability of planetary rings?
Centripetal acceleration explains how particles in planetary rings maintain their orbits. Each particle experiences an acceleration towards the planet, allowing it to follow a nearly circular path. This concept is crucial for modeling the long-term stability of ring systems.
25. How does the derivation of centripetal acceleration help in understanding the motion of charged particles in magnetic fields?
The derivation helps explain why charged particles move in circular paths in uniform magnetic fields. The magnetic force acts perpendicular to the velocity, providing the centripetal acceleration necessary for circular motion.
26. What is the role of centripetal acceleration in planetary motion?
Centripetal acceleration explains how planets maintain their orbital paths around the Sun. The Sun's gravity provides the centripetal force, resulting in an acceleration that continuously changes the planet's velocity direction, keeping it in orbit.
27. How does the concept of centripetal acceleration apply to electrons in an atom?
In the Bohr model of the atom, electrons are thought to orbit the nucleus in circular paths. The electrostatic attraction between the electron and the nucleus provides the centripetal force, resulting in the centripetal acceleration necessary for this orbital motion.
28. How does understanding centripetal acceleration help in analyzing the stability of spinning objects?
Understanding centripetal acceleration is crucial for analyzing spinning object stability. It helps explain why increasing angular velocity can stabilize rotation (like a gyroscope) by requiring a larger force to change the object's plane of rotation.
29. How does centripetal acceleration relate to the concept of angular momentum?
While centripetal acceleration doesn't directly change angular momentum, it's essential for maintaining the circular path that gives rise to angular momentum. The force causing centripetal acceleration ensures the conservation of angular momentum in circular motion.
30. What is the difference between centripetal acceleration and centrifugal force?
Centripetal acceleration is a real acceleration towards the center of circular motion. Centrifugal force is a fictitious force that appears to act outward in a rotating reference frame, but it's not a real force in an inertial frame of reference.
31. What role does mass play in centripetal acceleration?
Mass does not directly affect centripetal acceleration. The acceleration depends only on velocity and radius. However, mass is important when considering the force required to produce this acceleration (F = ma).
32. How does centripetal acceleration relate to Newton's First Law of Motion?
Centripetal acceleration explains why objects in circular motion don't follow Newton's First Law, which states that objects in motion tend to stay in motion in a straight line. The centripetal force causing this acceleration prevents the object from moving in a straight line.
33. How does gravity provide centripetal acceleration for orbiting bodies?
For orbiting bodies, gravity acts as the centripetal force, providing the necessary centripetal acceleration. The gravitational force always points towards the center of the orbit, continuously changing the direction of the orbiting body's velocity.
34. How does centripetal acceleration relate to the concept of inertia?
Centripetal acceleration overcomes an object's inertia, which would otherwise cause it to move in a straight line. The acceleration constantly changes the object's direction, counteracting its tendency to maintain its current state of motion as described by Newton's First Law.
35. What is centripetal acceleration?
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, always directed towards the center of the circle. It causes the object's velocity to change in direction, but not in magnitude, maintaining circular motion.
36. Why is centripetal acceleration necessary for circular motion?
Centripetal acceleration is necessary for circular motion because it continuously changes the direction of the object's velocity vector, keeping it moving in a circular path. Without this acceleration, the object would move in a straight line tangent to the circle.
37. How is centripetal acceleration different from tangential acceleration?
Centripetal acceleration changes only the direction of velocity, while tangential acceleration changes the magnitude of velocity. Centripetal acceleration is always perpendicular to the velocity, pointing towards the center of the circle, whereas tangential acceleration is parallel to the velocity.
38. Why doesn't centripetal acceleration change the speed of an object in circular motion?
Centripetal acceleration is always perpendicular to the velocity vector. Since it acts at right angles to the motion, it only changes the direction of velocity, not its magnitude (speed).
39. Why is the derivation of centripetal acceleration important in physics?
The derivation of centripetal acceleration is crucial because it explains the mechanics of circular motion, which is fundamental in many physical systems, from planetary orbits to particle accelerators. It bridges kinematics and dynamics in curvilinear motion.
40. What happens to centripetal acceleration if an object's speed in circular motion doubles?
If the speed doubles while the radius remains constant, the centripetal acceleration increases by a factor of four. This is because centripetal acceleration is proportional to the square of velocity (a = v²/r).
41. How does the concept of centripetal acceleration apply to satellites orbiting Earth?
Satellites experience centripetal acceleration due to Earth's gravity. This acceleration keeps them in orbit by constantly changing their velocity's direction. The satellite's speed and altitude are carefully calculated to balance this acceleration with Earth's gravitational pull.
42. How does the concept of centripetal acceleration apply to banked curves on roads?
On banked curves, the road's tilt provides a component of the normal force that acts as the centripetal force, producing the necessary centripetal acceleration. This allows vehicles to navigate turns at higher speeds without relying solely on friction.
43. What is the relationship between centripetal acceleration and the tension in a string for an object in circular motion?
For an object swung in a circle on a string, the tension in the string provides the centripetal force necessary for circular motion. The tension is directly related to the centripetal acceleration by T = ma, where 'm' is the mass of the object.
44. How does air resistance affect centripetal acceleration in real-world scenarios?
Air resistance can reduce the tangential velocity of an object in circular motion, which in turn reduces the centripetal acceleration required to maintain the circular path. In many practical scenarios, this effect is often negligible for short durations.
45. What role does centripetal acceleration play in the formation of galaxies?
Centripetal acceleration is crucial in galaxy formation and structure. It explains how stars and gas maintain their orbital paths around the galactic center, with gravity providing the necessary centripetal force for this large-scale circular motion.
46. How does the concept of centripetal acceleration apply to the motion of blood in curved arteries?
Blood flowing through curved arteries experiences centripetal acceleration. This concept is important in understanding blood flow dynamics, pressure distributions in vessels, and potential areas of plaque buildup in the cardiovascular system.
47. What is the relationship between centripetal acceleration and the Coriolis effect?
While centripetal acceleration is real, the Coriolis effect is an apparent force in rotating reference frames. Both are important in understanding motion on a rotating Earth, but they describe different aspects of motion relative to rotating systems.
48. How does the derivation of centripetal acceleration help in understanding the concept of gravitational slingshot maneuvers in space missions?
The derivation helps explain how a spacecraft can gain speed by passing close to a planet. The spacecraft's path curves due to the planet's gravity (providing centripetal acceleration), and it can exit the encounter with increased velocity relative to the Sun.
49. Why is it important to consider centripetal acceleration when designing centrifugal pumps?
In centrifugal pumps, understanding centripetal acceleration is crucial for designing the impeller shape and rotation speed. It helps engineers calculate the forces on the fluid, optimize energy transfer, and predict the pump's performance under various conditions.
50. How does the derivation of centripetal acceleration account for changes in the radius of circular motion?
The derivation assumes a constant radius. For motion where the radius changes, like spiral motion, the acceleration would have both a centripetal component and a radial component, requiring a more complex analysis.
51. What is the significance of the negative sign in the vector form of centripetal acceleration?
The negative sign in the vector form of centripetal acceleration (a = -ω²r) indicates that the acceleration vector points in the opposite direction of the position vector, i.e., towards the center of the circle.
52. Why is the centripetal acceleration vector always perpendicular to the velocity vector in circular motion?
The centripetal acceleration vector is always perpendicular to the velocity vector because it needs to change only the direction of motion, not the speed. This perpendicular relationship ensures that the object maintains a constant speed while following a circular path.
53. How does the derivation of centripetal acceleration relate to the concept of instantaneous velocity in circular motion?
The derivation of centripetal acceleration uses the concept of instantaneous velocity, which is tangent to the circular path at any point. The rate of change of this velocity vector's direction gives rise to the centripetal acceleration.
54. How does the derivation of centripetal acceleration contribute to our understanding of the equivalence principle in general relativity?
The derivation of centripetal acceleration helps illustrate the equivalence between gravitational acceleration and acceleration due to other forces. This concept is fundamental to Einstein's equivalence principle, which states that the effects of gravity are indistinguishable from those of acceleration in a small region of spacetime.

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