When an object travels in a circular path, then the force that keeps it fixed in its path is the centripetal force. Moreover, in this topic, we will discuss the definition example and formula of it.

**Centripetal Force**

Marry-go-round is the perfect example of centripetal force as when they operate people hold the bar and as the speed of the merry-go-round increases the harder it becomes to hold on to the bar. Also, this creates equilibrium between two forces one real and one apparent- in order to stay in that circular path.

Furthermore, merry-go-round helps to explain how force helps to keep an object moving in a circular path. Moreover, your body wants to fly off the merry-go-round in a straight line but your hands exert an opposite force to keep you on the ride.

Besides, the propensity of your body to fly off the merry-go-round is the centrifugal force. In addition, it is an apparent force and not a real one. And the force that your hands use to stay on the ride is the centripetal force.

**Definition**

It refers to the force that keeps an object moving in its circular path and helps it to stay on the path. Also, its value is determined by three factors:

1- The velocity of the object when it follows the circular path.

2- The distance of object from the center of path

3- Mass of the object.

Besides, centrifugal force is the tendency of objects to leave the circular path and fly off in a straight line. Sometimes people mistakenly assume centripetal force to be a centrifugal force.

Moreover, the velocity of the object is constant and perpendicular to the line running from the object to the center of the circle and is called tangential velocity.

**Calculation of Centripetal Force**

It is easy to calculate centripetal force if you know the mass of the object (m), the distance of the object or radius (r) from the center, and the tangential velocity, v. Moreover, the equation is based on the metric system and centripetal force is written as f. Besides, we use Newtons to measure it and one Newton is approximately 0.225 lb (102.05 g).

\(F_c\) = \(\frac{mv²}{r}\)[Newtons, N]

When an object travels in a circular path, then the force that keeps it fixed in its path is the centripetal force. Moreover, in this topic, we will discuss the definition example and formula of it.

Most noteworthy, there are some interesting things about this equation. The tangential velocity is square and if you double it then quadruples the centripetal force. Moreover, the r shows as the denominator, so the magnitude of centrifugal force decreases as the object gets further away from the center.

**Example**

Let’s take an example. Assume that in the merry-go-round situation Mr. X is standing at the edge of the ride holding to the bar and he weighs 70 pounds. Besides, the diameter of the merry-go-round is 3 meters and the ride complete one revolution in 4 seconds. Then what will be the centripetal force that Mr. X must exert to stay on the ride?

Firstly, the circumference of the ride is the diameter of the merry-go-round multiplied by pi (π). Besides, this calculation gives us 9.4 meters around the perimeter of the ride. Moreover, if Mr. X is traveling 9.4 meters per 4 seconds then the tangential velocity is 9.4 / 4 = 2.35 meter per second.

Now let’s find the radius by dividing the diameter by 2 that is 1.5 meters. Now we need to covert Mr. X’s weight into the mass that is 0.454 kg for one pound. In this way, his mass will be 70 × 0.454 = 32 kg. Now put the values in the formula.

\(F_c\) = \(\frac{mv^2}{r}\) =\( \frac{(32)(2.35)^2}{1.5}\)\) = 118 N

So, Mr. X exerts a centripetal force of 118 Newtons to stay on the ride.

**Solved Question for You**

Question. What is Centrifugal force?

A. The tendency to fly off from a circular path

B. The tendency to stay on the circular path

C. The tendency to resist change

D. None of the above.

Answer. The correct answer is option A.

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