Any force can be a centripetal force. Centripetal force is the umbrella term given to any force that is acting perpendicular to the motion of an object and thus causing it to go in a circle. Keep the following in mind when solving centripetal force type problems:

- The speed of the object remains constant. The centripetal force is changing the direction but not the speed of the object.
- Although the object 'feels' an outward pull, this is not a true force, but merely the objects inertia. Remember, Newton's first law maintains that the natural state of an object is to go in a straight line at constant speed. Thus, when you make a right turn in your car and the basketball in the back seat flies to the left, that is because the car is moving right and the basketball is maintaining it's position and thus from your point of view moves to the left. Your point of view in this case is different from reality because you are in a rotating reference frame.

Key Equations

Centripetal Force\begin{align*} F_C = \frac{mv^2}{r} \begin{cases}
m & \text{mass (in kilograms, kg)}\\
v & \text{speed (in meters per second, m/s}\text{)}\\
r & \text{radius of circle}
\end{cases}\end{align*}

#### Applications

- To find the maximum speed that a car can take a corner on a flat road without skidding out, set the force of friction equal to the centripetal force.

- To find the tension in the rope of a swinging pendulum, remember that it is the
*sum*of the tension and gravity that produces a net upward centripetal force. A common mistake is just setting the centripetal force equal to the tension.

- To find the speed of a planet or satellite in an orbit, set the force of gravity equal to the centripetal force.

- Banked turns (the road is not flat, but rather at an angle) allow for a greater speed before skidding out. In this case the normal force aids the force of friction in creating a larger centripetal force than that which can be obtained by friction alone on a flat road. It’s not as simple as adding them together, one must consider the components of these two forces in the direction to the center of the circle.

### Simulation

### Review

- A 700kg car makes a turn going at 30 m/s with radius of curvature of 120m. What is the force of friction between the car’s tires and the road?
- An object of mass \begin{align*}10 \;\mathrm{kg}\end{align*} is in a circular orbit of radius \begin{align*}10 \;\mathrm{m}\end{align*} at a velocity of \begin{align*}10 \;\mathrm{m/s}\end{align*}.
- Calculate the centripetal force (in \begin{align*}N\end{align*}) required to maintain this orbit.
- What is the acceleration of this object?

- Suppose you are spinning a child around in a circle by her arms. The radius of her orbit around you is \begin{align*}1\end{align*} meter. Her speed is \begin{align*}1 \;\mathrm{m/s}\end{align*}. Her mass is \begin{align*}25 \;\mathrm{kg}\end{align*}.
- What is the magnitude and direction of tension in your arms?
- In her arms?

- A racecar is traveling at a speed of \begin{align*}80.0 \;\mathrm{m/s}\end{align*} on a circular racetrack of radius \begin{align*}450 \;\mathrm{m}\end{align*}.
- What is its centripetal acceleration in \begin{align*}\;\mathrm{m/s}^2\end{align*}?
- What is the centripetal force on the racecar if its mass is \begin{align*}500 \;\mathrm{kg}\end{align*}?
- What provides the necessary centripetal force in this case?

- The radius of the Earth is \begin{align*}6380 \;\mathrm{km}\end{align*}. Calculate the velocity of a person standing at the equator due to the Earth’s 24 hour rotation. Calculate the centripetal acceleration of this person and express it as a fraction of the acceleration g due to gravity. Is there any danger of “flying off”?
- Neutron stars are the corpses of stars left over after supernova explosions. They are the size of a small city, but can spin several times per second. (Try to imagine this in your head.) Consider a neutron star of radius \begin{align*}10\;\mathrm{km}\end{align*} that spins with a period of \begin{align*}0.8\end{align*} seconds. Imagine a person is standing at the equator of this neutron star.
- Calculate the centripetal acceleration of this person and express it as a multiple of the acceleration \begin{align*}g\end{align*} due to gravity (on Earth).
- Now, find the minimum acceleration due to gravity that the neutron star must have in order to keep the person from flying off.

### Review (Answers)

- 5250 N
- a. \begin{align*}100 \;\mathrm{N}\end{align*} b. \begin{align*}10 \;\mathrm{m/s}^2\end{align*}
- a. \begin{align*}25 \;\mathrm{N}\end{align*} towards her b. \begin{align*}25 \;\mathrm{N}\end{align*} towards you
- a. \begin{align*}14.2 \;\mathrm{m/s}^2\end{align*} b. \begin{align*}7.1 \times 10^3 \;\mathrm{N}\end{align*} c. friction between the tires and the road
- \begin{align*}.0034\mathrm{g}\end{align*}
- a. \begin{align*}6.3 \times 10^4g\;\mathrm{m/s}^2\end{align*} b. The same as a.