Chapter 11: Basic Physics SECircular Motion
The Big Idea
In the absence of a net force, objects move in a straight line. If they turn — that is, if their velocity changes, even only in direction — there must be an applied force. Forces which cause objects to turn around continuously in a circle are known as centripetal forces. When an object moves in a circle its velocity at any particular instant points in a direction tangent to the circle. The acceleration points towards the center of the circle, and so does the force acting on it. This is only natural, when you think about it — if you feel a force pushing you towards your left as you walk forward, you will walk in a circle, always turning left.
Key Equations
 \begin{align*}a_C = \frac{v^2}{r}\end{align*} ; the acceleration of an object moving at constant speed and going in a circle is equal to the square of its speed divided by the radius of the circle.
 \begin{align*}F_C = ma_C = \frac{mv^2}{r}\end{align*} ; if an object of mass \begin{align*}m\end{align*} moves in a circle of radius \begin{align*}r\end{align*} at speed \begin{align*}v\end{align*}, the force acting on it must be equal to this value. The direction of this force is always pointed towards the center of the circle.
Key Concepts
 Centripetal Force is an umbrella term that refers to any force that makes something move in a circle. Examples of forces that can be centripetal forces are as follows: Gravity (the gravity between the Earth and Sun causes the Earth to go in a circle), Friction (the friction force between a car’s tires and the road causes the car to turn), Tension (when swinging a bucket around in a circle by holding onto the string, the tension of the string is the centripetal force)
 An object moving in a circle has an instantaneous velocity tangential to the circle of its path. The force and acceleration point to the center of the circle.
 Net force and acceleration always have the same direction.
 An orbital period, \begin{align*}T\end{align*}, is the time it takes to make one complete rotation.
 If a particle travels a distance \begin{align*}2 \pi r\end{align*} in an amount of time \begin{align*}T\end{align*}, then its speed is distance over time or \begin{align*}\frac{2 \pi r}{T}\end{align*}.
 The centripetal force can by comprised of more than one force. For example, with roller coasters, the centripetal force at the top of a loop is the addition of the force of gravity and the normal force, both of which are acting on the rollercoaster and both pointing down from the top and, thus, towards the center of the circle. Note at the bottom of the loop, one would subtract the force of gravity from the normal force to find the centripetal force.
 Centripetal acceleration is just the acceleration provided by centripetal forces and can be found by dividing the centripetal force by the objects mass.
Key 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.
 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.
 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, realize that the force of gravity is the centripetal force. So set the force of gravity equal to \begin{align*}\frac{mv^2}{r}\end{align*}, where \begin{align*}v\end{align*} is the speed of the planet, \begin{align*}m\end{align*} is the mass of the planet or satellite and \begin{align*}r\end{align*} is the distance of the planet to the sun or the satellite to the Earth.
 Some data needed for the problems:
 The radius of Earth is \begin{align*}6.38 \times 10^6 \ m\end{align*}
 The mass of Earth is about \begin{align*}6.0 \times 10^{24} \ kg\end{align*}
 The mass of Sun is about \begin{align*}2.0 \times 10^{24} \ kg\end{align*}
 The EarthSun distance is about \begin{align*}1.5 \times 10^{11} \ m\end{align*}
 The EarthMoon distance is about \begin{align*}3.84 \times 10^8 \ m\end{align*}
Centripetal Forces Problem Set
 When you make a right turn at constant speed in your car what is the force that causes you (not the car) to change the direction of yourvelocity? Choose the best possible answer.
 Friction between your butt and the seat
 Inertia
 Air resistance
 Tension
 All of the above
 None of the above
 You buy new tires for your car in order to take turns a little faster. The new tires double the force of friction between your tires and the road. With the old tires you could take a particular turn at a speed \begin{align*}v_o\end{align*}. What is the maximum speed you can now take the turn without skidding out?
 \begin{align*}4v_o\end{align*}
 \begin{align*}2v_o\end{align*}
 \begin{align*}v_o\end{align*}
 \begin{align*}\sqrt{2}v_o\end{align*}
 Not enough information given
 A pendulum consisting of a rope with a ball attached at the end is swinging back and forth. As it swings downward to the right the ball is released at its lowest point. Decide which way the ball attached at the end of the string will go at the moment it is released.
 Straight upwards
 Straight downwards
 Directly right
 Directly left
 It will stop
 A ball is spiraling outward in the tube shown to the right. Which way will the ball go after it leaves the tube?
 Towards the top of the page
 Towards the bottom of the page
 Continue spiraling outward in the clockwise direction
 Continue in a circle with the radius equal to that of the spiral as it leaves the tube
 None of the above
 Explain using Newton’s Second Law why an object moving in a circle must experience a net force towards the center of the circle.
 Explain the difference between an inertial and a noninertial frame of reference, and give an example of each.
 In the book and movie “2001: A Space Odyssey” (Google it) Arthur C. Clarke described a large space habitat that would rotate to simulate gravity. Explain thoroughly how this would work. Include a diagram in your answer, and show where the inhabitants would walk. Why would it be physiologically better to make this habitat large? What force is the “gravity”? (also see optional problem section, #16, for an analysis of this space habitat)
 In the carnival ride called the “Gravitron” riders are spun in a cylindrical room at increasing speed. When a certain speed is reached the floor drops away and the riders are plastered against the wall. Draw a FBD for a rider in the Gravitron and identify all the forces acting on them. What force holds them in? What force holds them up? What force pulls them out? Is this an inertial or noninertial frame of reference?
 Many turns on roads are “banked” – instead of being flat, the road is angled to allow the cars to turn easier. Why? Explain, and use a diagram to assist with your explanation.
 An object of mass 10 kg is in a circular orbit of radius 10 m at a velocity of 10 m/s.
 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 1 meter. Her speed is 1 m/s. Her mass is 25 kg.
 What is the tension in your arms?
 In her arms?
 A racecar is traveling at a speed of 80.0 m/s on a circular racetrack of radius 450 m.
 What is its centripetal acceleration in \begin{align*}m/s^2\end{align*}?
 What is the centripetal force on the racecar if its mass is 500 kg?
 What provides the necessary centripetal force in this case?
 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. Is there any danger of “flying off”? Explain.
 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 10 km that spins with a period of 0.8 seconds. Imagine a person is standing at the equator of this neutron star.
 Calculate the centripetal acceleration of this person and compare it to 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.
 A student comes up to you and says, “I can visualize the force of tension, the force of friction, and the other forces, but I can’t visualize centripetal force.” As you know, a centripetal force must be provided by tension, friction, or some other “familiar” force. Write a two or three sentence explanation, in your own words, to help the confused student.
 Calculate the centripetal acceleration of the Earth around the Sun.
 You are speeding around a turn of radius 30.0 m at a constant speed of 15.0 m/s. Assume the mass of the car is 800 kg.
 What is the minimum force of friction between your car tires and the road necessary for you to retain control?
 Even if the road is terribly icy, you will still move in a circle because you are slamming into the walls. What centripetal forces (in the form of a Normal Force) must the walls exert on you. Assume you do not lose speed during the turn.
Answers:

 100 N
 \begin{align*}10 \ m/s^2\end{align*}

 25 N towards her
 25 N towards you

 \begin{align*}14.2 \ m/s^2\end{align*}
 \begin{align*}7.1 \times 10^3 \ N\end{align*}
 friction between the tires and the road
 \begin{align*}v = 464 \ m/s; a = 0.034 \ m/s^2\end{align*}

 \begin{align*}6.2 \times 10^5 \ m/s^2\end{align*}
 The same as a.
 discuss in class
 \begin{align*}.006 \ m/s^2\end{align*}

 6000 N
 same as in a, if assume no friction
OPTIONAL PROBLEM
 A space station was established far from the gravitational field of Earth. Extended stays in zero gravity are not healthy for human beings. Thus, for the comfort of the astronauts, the station is rotated so that the astronauts feel there is an internal gravity. The rotation speed is such that the apparent acceleration of gravity is \begin{align*}9.8 \ m/s^2\end{align*}. The direction of rotation is counterclockwise.
 If the radius of the station is 80 m, what is its rotational speed, \begin{align*}v\end{align*}?
 Draw vectors representing the astronaut’s velocity and acceleration.
 Draw a free body diagram for the astronaut.
 Is the astronaut exerting a force on the space station? If so, calculate its magnitude. Her mass \begin{align*}m = 65 \ kg\end{align*}.
 The astronaut drops a ball, which appears to accelerate to the ‘floor’, (see picture) at \begin{align*}9.8 \ m/s^2\end{align*}.
 Draw the velocity and acceleration vectors for the ball while it is in the air.
 What force(s) are acting on the ball while it is in the air?
 Draw the acceleration and velocity vectors after the ball hits the floor and comes to rest.
 What force(s) act on the ball after it hits the ground?
Answers:

 \begin{align*}v = 28m/s\end{align*}
 \begin{align*}v\end{align*}down, \begin{align*}a\end{align*}right
 \begin{align*}f\end{align*}right
 Yes, 640 N