You have landed on the right page to learn Centripetal Force Formula. For instance, you may have observed that the slope is provided at the turning of the roads. Why are slopes provided at turns? The answer is Centripetal Force. When a vehicle makes a turn, it will move in the arc of a circle and will require centripetal force. When the vehicle passes through the inclined road turn, a component of weight and friction provides the required centripetal force. We can see many examples of centripetal force in our daily lives.

You will experience this force when you swing a stone tied with a string or when you move around a circular track while sitting in a car or when an aeroplane banks into a turn. The centripetal force is not a fundamental force; it is the net force that causes an object to move in a circular path. The gravitational force keeping a satellite in orbit is also an example of centripetal force.

In this article, we will discuss how centripetal force works and what is the concept behind the centripetal force. We will also discuss centripetal acceleration and its expression to calculate the centripetal force.

Centripetal Force

As we know that the acceleration of a particle in circular motion has two components, tangential and radial (or centripetal). So, normally we resolve the forces acting on the particle in two directions, one is tangential, and the other is radial. In tangential direction, the net force on the particle is \(m{a_t}\) and in radial direction, net force is \(m{a_c}\). But, in a uniform circular motion, tangential acceleration \(\left( {{a_t}} \right)\) is zero. Hence, net force in the tangential direction is zero, and in the radial direction, the net force is given as:

\({F_{net}} = m{a_c} = \frac{{m{v^2}}}{r}\) and \(F = mr{\omega ^2}\)

\({a_c} = \frac{{{v^2}}}{r} = r{\omega ^2}\)

This net force (towards the centre) is called centripetal force, as shown in the above figure. This much force is required for the particle to rotate in a circle (as it is accelerated due to a change in the direction of velocity). The real forces acting on the particle, such as friction force, weight, normal reaction, tension, etc., provide this centripetal force.

Centripetal Force Formula

In a uniform circular motion, there is always an associated acceleration as the direction of the velocity changes constantly, even though the magnitude of the velocity might be constant. When you take a turn on a circular track in your car (moving at a constant speed), then you will experience this acceleration. As you and the car change direction, you will notice a sideways acceleration. This acceleration will be more noticeable when the curve will be sharper, and your speed will be greater. Now, let us find the direction and magnitude of that acceleration.

Centripetal Acceleration

Let us consider an object of mass \((m)\) moving along the circumference of a circle of radius \((r)\) with uniform speed \((v)\) as shown in the figure given below. At the two points along the path, the direction of the instantaneous velocity is shown in this figure. Consider the velocity at two moments during the circular path, where \(\overrightarrow {{v_1}} \) represents the initial velocity and \(\overrightarrow {{v_2}} \) represents the velocity after a short time interval. The direction of the acceleration is in the direction of the changing velocity, pointing directly toward the centre of rotation, as shown in the figure with the vector diagram. Then, we can say that the acceleration of an object moving in a uniform circular motion. The triangle formed by the radii \((r)\) and \(AB\) and the triangle formed by the velocity vectors is similar. Both the triangles \(AOB\) and \(PQS\) are isosceles (two equal sides). We can write the two equal sides of the velocity vector triangle as the speeds \(\left| {{v_1}} \right| = \left| {{v_2}} \right| = v.\)

So from the above figure, we can write :

\(\overrightarrow {PQ} + \overrightarrow {QS} = \overrightarrow {PS} \)

\( – \overrightarrow {{v_1}} + \overrightarrow {{v_2}} = \overrightarrow {\Delta v} \)

\(\overrightarrow {\Delta v} = \overrightarrow {{v_2}} – \overrightarrow {{v_1}} \)

And, also the triangle \(PQS\) and \(AOB\) are similar, Therefore,

\(\frac{{\Delta v}}{{AB}} = \frac{v}{r}……(1)\)

And,   \(AB = {\mathop{\rm arc}\nolimits} AB = v\Delta t\)        (for very small \(∆t)……..(2)\) 

Using equations \((1)\) and \((2),\) we have

\(\frac{{\Delta v}}{{v\Delta t}} = \frac{v}{r}\)

\( \Rightarrow \frac{{\Delta v}}{{\Delta t}} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{v^2}}}{r}\quad \ldots \left( 3 \right)\)

In terms of angular velocity \(\left( \omega \right),\) we can express \(a_c\) by substituting \(v=rω,\) in the above equation \((3),\) we get

\({a_c} = \frac{{{{(r\omega )}^2}}}{r} = r{\omega ^2} \cdots \left( 4 \right)\)

Then, the magnitude of the centripetal acceleration can be expressed by using either of two equations \((3)\) or by \((4)\)

\({a_c} = \frac{{{\nu ^2}}}{r}\) and  \({a_c} = r{\omega ^2}\)

Direction: The direction of ac is toward the centre.

Centripetal Force Formula

If we know the acceleration of the body, Then by Newton’s law, we can write,

Net force, \(F=ma,\)

This force is the centripetal force that is equal to the mass of the object multiplied by the centripetal acceleration\((a_c).\)

i.e

\(F = \frac{{m{v^2}}}{r}\) and  \(F = mr{\omega ^2}\)

Examples of Centripetal Force in Daily Life

As per the centripetal force definition, we can say that it is the force that pulls or pushes an object toward the centre of a circle as it travels, causing circular motion. Below are some of the examples of the Centripetal Force.

Applications  of Centripetal Force

1. Twirling a lasso or spinning a ball on a string: Here, the force of tension on the rope provides the centripetal force that pulls the object in toward the centre.
2. Turning of a car on a circular track: In this case, centripetal force is provided by the frictional force between the ground and the wheels. On high-speed highways, a slope is provided at the turns. In this case, a component of weight contributes to the required centripetal force.
3. Moving through a loop on a roller coaster: The force is provided by the Normal Force as the seat or wall pushes you toward the centre.
4. Planets orbiting around the Sun: In this case, Centripetal Force is provided by the force of Gravity between sun and planet.

Sample Problems on Centripetal Force Formula

Q.1.Find the normal constant force by the sidewall of the groove when a small block of mass \({\rm{100}}\,{\rm{g}}\) moves with uniform speed in a horizontal circular groove if the block takes \({\rm{2}}{\rm{.0}}\,{\rm{s}}\) to complete one round, and the radius of the vertical sidewalls is \({\rm{25}}\,{\rm{cm}}\)?
Ans: The speed of the block is
\(v = \frac{{2\pi \times 25}}{{2.0 \times 100}} = 0.785\;{\rm{m}}\,{{\rm{s}}^{ – 1}}\)
The acceleration of the block is
\({a_c} = \frac{{{v^2}}}{r} = \frac{{{{0.785}^2}}}{{0.25}} = 2.4649\;{\rm{m}}\,{{\rm{s}}^{ – 2}}\left( {{\rm{towards}}\,{\rm{the}}\,{\rm{centre}}} \right)\)
The normal contact force due to the side walls is the only force in this direction. Thus, from Newton’s second law, this force will be
\(F = m{a_c}\)
\( \Rightarrow F = (0.1)(2.464)\;{\rm{N}}\)
\( \Rightarrow F = 0.246\;{\rm{N}}\)

Q.2. Find the force exerted by the air on a fighter plane of mass \({\rm{1600}}\,{\rm{kg}}\) at the lowest point that is pulling out for a dive at a speed of \({\rm{900}}\,{\rm{km}}\,{{\rm{h}}^{{\rm{ – 1}}}}{\rm{.}}\) Consider the path to be a vertical circle whose radius is given as \({\rm{2000}}\,{\rm{m}}{\rm{.}}\) Take, \({\rm{g}} = 9.8\;{\rm{m}}\,{{\rm{s}}^{ – 2}}\)
Ans: At the lowest point, the acceleration is vertically upward, and its magnitude is given as, \({a_c} = \frac{{{v^2}}}{r}.\)
The forces on the plane are:
(a) weight \(Mg\) downward and
(b) Force \(F\) by the air upward.
Given,
\(v = 900\;{\rm{km}}\;{{\rm{h}}^{ – 1}} = \frac{{900 \times {{10}^3}}}{{3600}}\;{\rm{m}}\;{{\rm{s}}^{ – 1}} = 250\;{\rm{m}}\;{{\rm{s}}^{ – 1}}\)
From Newton’s second law, we have
\(F – Mg = \frac{{M{v^2}}}{r}\)
\( \Rightarrow F = M\left( {g + \frac{{{v^2}}}{r}} \right)\)
\( \Rightarrow F = 1600\left( {9.8 + \frac{{{{250}^2}}}{{2000}}} \right){\rm{N}}\)
\( \Rightarrow F = 6.56 \times {10^5}\;{\rm{N}}\,{\rm{(Upward)}}\)

Summary

When an object of mass \((m)\) revolving in a circular motion of radius \((r),\) the object is in accelerating motion. The radial component of the acceleration, called centripetal acceleration is given by, \({a_c} = \frac{{{v^2}}}{r}.\) Its direction is towards the centre of the circular orbit. In a uniform circular motion, the magnitude of the velocity \((v)\) is constant and only direction is changing. Since velocity is a vector quantity, it will keep on changing. This change in velocity is caused by centripetal force, which is directed towards the centre and given by, 

\(F = m{a_c} = \frac{{m{v^2}}}{r} = mr{\omega ^2}.\)

A centripetal force is also known as a radial force.

FAQs on Centripetal Force Formula

Q.1. How is centripetal force calculated?
Ans: According to Newton’s second law of motion, the net force is mass times acceleration: \({F_{net}} = ma.\) For uniform circular motion, the acceleration is the centripetal acceleration \(\left( {{a_c}} \right).\) Thus, the magnitude of centripetal force \(\left( {{F_c}} \right)\) is given as \({F_c} = m{a_c}.\)

Q.2. Define centripetal force and write its formula?
Ans: Centripetal force is the force that acts on the body to keep it moving in a curved path. It is directed inward towards the centre of rotation. Expression for the centripetal formula is given as :
\(F = \frac{{M{v^2}}}{r} = mr{\omega ^2}\)

Q.3. What are 3 examples of centripetal force?
Ans: When viewed from a rotating frame of reference, centrifugal force acts on every object moving in a circular path. Centripetal Force examples are:
(i). Spinning a ball on a string
(ii). The motion of planets around the sun
(iii). Loops in a roller coaster.

Q.4. Write the difference between centripetal force and centrifugal force?
Ans: The force that is required to keep an object moving in a curved path and that is directed inward towards the centre of rotation is known as a centripetal force, while the apparent force that is felt by an object moving in a curved path that acts outwardly away from the centre of rotation is known as centrifugal force.

Q.5. Why is the centripetal force necessary for a circular motion?
Ans: To move in a circular motion, we need to apply a centripetal force that changes the direction of the velocity. Otherwise, according to Newton’s First Law, if there was no net force, the object would move straight with constant velocity.

We hope you find this article on ‘Centripetal Force Formula‘ helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them.