Learn what centripetal forces are and how to calculate them.

What is a centripetal force?

A centripetal force is a net force that acts on an object to keep it moving along a circular path.
In our article on centripetal acceleration, we learned that any object traveling along a circular path of radius r with velocity v experiences an acceleration directed toward the center of its path,
a, equals, start fraction, v, start superscript, 2, end superscript, divided by, r, end fraction.
However, we should discuss how the object came to be moving along the circular path in the first place. Newton’s 1ˢᵗ law tells us that an object will continue moving along a straight path unless acted on by an external force. The external force here is the centripetal force.
It is important to understand that the centripetal force is not a fundamental force, but just a label given to the net force which causes an object to move in a circular path. The tension force in the string of a swinging tethered ball and the gravitational force keeping a satellite in orbit are both examples of centripetal forces. Multiple individual forces can even be involved as long as they add up (by vector addition) to give a net force towards the center of the circular path.
Starting with Newton's 2ⁿᵈ law :
a, equals, start fraction, F, divided by, m, end fraction
and then equating this to the centripetal acceleration,
start fraction, v, start superscript, 2, end superscript, divided by, r, end fraction, equals, start fraction, F, divided by, m, end fraction
We can show that the centripetal force F, start subscript, C, end subscript has magnitude
F, start subscript, c, end subscript, equals, start fraction, m, v, start superscript, 2, end superscript, divided by, r, end fraction
and is always directed towards the center of the circular path. Equivalently, if omega is the angular velocity then because v, equals, r, omega,
F, start subscript, c, end subscript, equals, m, r, omega, start superscript, 2, end superscript

Tethered ball

One apparatus that clearly illustrates the centripetal force consists of a tethered mass (m, start subscript, 1, end subscript) swung in a horizontal circle by a lightweight string which passes through a vertical tube to a counterweight (m, start subscript, 2, end subscript) as shown in Figure 1.
Figure 1: Demonstration of a centripetal force provided by a mass m2 holding a spinning tethered ball.
Figure 1: Demonstration of a centripetal force provided by a mass m, start subscript, 2, end subscript holding a spinning tethered ball.
Exercise 1: If m, start subscript, 1, end subscript is a 1 kg1~\mathrm{kg} mass spinning in a circle of radius 1 m1~\mathrm{m} and m2=4 kgm_2 = 4~\mathrm{kg} what is the angular velocity assuming neither mass is moving vertically and there is minimal friction between the string and tube?
When the string passes through the tube it redirects the force due to gravity acting on m, start subscript, 2, end subscript to the horizontal plane. This is the centripetal force which allows m, start subscript, 1, end subscript to rotate in a circle. Neither mass is moving vertically so the centripetal force must be exactly balanced.
m, start subscript, 1, end subscript, r, omega, start superscript, 2, end superscript, equals, m, start subscript, 2, end subscript, g
Which can be rearranged and solved for the rotational velocity,
ω=m2gm1r=4 kg9.81 m/s21 kg1 m=6.26 rad/s\begin{aligned} \omega &= \sqrt{\frac{m_2 g}{m_1 r}} \\ &= \sqrt{\frac{4~\mathrm{kg}\cdot 9.81~\mathrm{m/s^2}}{1~\mathrm{kg}\cdot 1~\mathrm{m}}} \\ &= 6.26~\mathrm{rad/s}\end{aligned}
or about 2 revolutions per second.

Wind turbine

Exercise 2a: A large wind turbine has blades of 35 m in length, each weighing 10,000 kg. The center of mass of a blade is located halfway along its length. If the turbine is rotating at 20 revolutions per minute, what tension force are the bolts under as they hold a spinning blade to the hub?
First converting the rotation speed to rad/s:
ω=20 rev2π rad/rev60 s2.1 rad/s\omega = \frac{20~\text{rev} \cdot 2\pi~\mathrm{rad/rev}}{60~\mathrm{s}} \simeq 2.1~\mathrm{rad/s}
And now finding the centripetal force:
FC=mω2r=(1104 kg)(2.1 rad/s)2(352 m)7.7105 N\begin{aligned}F_C &= m\omega^2 r \\ &= (1\cdot 10^4~\mathrm{kg})\cdot (2.1~\mathrm{rad/s})^2 \cdot \left( \frac{35}{2}~\mathrm{m} \right) \\ &\simeq 7.7\cdot 10^5~\mathrm{N} \end{aligned}
Note that here we are using the distance to the center of mass for the radius of rotation. We can do this because the center of mass is the point through which a uniform force on the blade acts. Though the internal tension required to keep the tip of the blade together will be higher than at the center of mass, overall it is balanced out by the smaller tension for the section near the hub.