Centripetal Force

5.3 Centripetal Force

» From Newton’s 2nd Law we know that whenever an object accelerates, there must be a net force acting on the object to create the acceleration.

» The net force causing centripetal acceleration is called centripetal force, and points towards the center of the circle. Fc

FcFc

»Newston’s 2nd Law say ΣF=ma, so…»ΣFradially= Fc = mac

»Fc = m(

The sum of the forces acting in the radial direction equals the mass times the centripetal acceleration

5.3 Centripetal Force

» “Centripetal force” is not a new or separate force!

» “Centripetal force” describes (or labels) the net force pointing toward the center of a circular path.

» It is the vector sum of all forces pointing along the radial direction.

» The centripetal force could be due to tension, or friction, or lift, etc. A FBD will help to identify which force(s) is(are) causing centripetal acceleration.

Example 1

» A model airplane has a mass of 0.90kg and moves at a constant speed on a circle that is parallel to the ground. The path of the airplane and its guideline lie in the same horizontal plane because the weight of the plane is balanced by the lift generated by its wings. Find the tension in the guideline for speed of 19m/s and 38m/s. The length of the guideline is 17m.

Speed = 19m/s; T=19NSpeed=38m/s; T=76N

»When a car moves without skidding around a curve, what provides the centripetal force to keep the car on the road?

» static friction between the road and the tires provides the centripetal force to keep the car on the road.

»When a car moves at a steady rate around an unbanked curve, the centripetal force keeping the car on the road comes from the static friction between the road and the tires.

» Some free body diagrams for a car going around a curve would look like this:

Example 2

» EX 7: At what maximum speed can a car safely negotiate a horizontal unbanked turn (radius = 51 m) in dry weather (coefficient of static friction = 0.95) and again in icy weather (coeff of static friction = 0.10)?

» Ff = μ N = μ mg

» Fc =

» Fc = Ff so = μ mg» Notice that the mass will cancel out of the equations above:

= μ mg» On dry road: v = = 21.6 m/s» On icy road: v = = 7.1 m/s

Example 3

» Kevin is playing on a merry-go-round in a park. He is hanging on to the railing at the edge of the merry-go-round, 1.5 m from the center.

» a) If Kevin has a mass of 65 kg, and the railing pulls him in with 260 N, how fast is he moving?

» b) How long will it take him to make one complete revolution around the merry-go-round? (Hint: How far does he have to travel to go around the circle?)

Example 4

» You go on the floor drop ride in a local amusement park. In the ride, you stand along the walls of a giant drum (5 m radius), which is rotated at fairly high speed. Once you are spinning at full speed, the floor is dropped out from underneath you, but you don’t fall with it. If the coefficient of friction between you and the wall is 0.8, how fast does the drum need to be spinning for you to stay up when the floor drops? Guess your mass as best you can to use in the calculation. (Hint: Think about what force makes you spin in a circle with the drum and how that is connected to the friction.) Answer: 7.83 m/s

Example 5

» When an airplane or stunt driver completes a loop de loop, they often vary their speed as they complete the circular turn (non-uniform circular motion). Nonetheless, they can maintain a constant speed and complete the turn. Let's assume uniform circular motion to examine the FBD at four locations on the loop. The propulsion and braking forces are omitted because they are not acting in the radial direction.

Example 5

» In every case, the sum of the Radial forces must = =

If the mass of the car is 1100kg, and the radius of the loop is 10m, what is the minimum speed that the car must maintain to make it around the loop?

Assignment

»P. 156 #13-17, 21, 23