Centripetal Force - PowerPoint Presentation

Slide1

Centripetal ForceSlide2

5.3 Centripetal Force

From

Newton’s 2

nd 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.

F

c

F

c

F

cSlide3

Newston’s

2

nd

Law say ΣF=ma, so…ΣFradially= F

c = ma

cFc = m(

The sum of the forces acting in the radial direction equals the mass times the centripetal accelerationSlide4

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.Slide5

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=19N

Speed=38m/s; T=76NSlide6

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.Slide7

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:Slide8

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 mgFc

=

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

 Slide9

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?)Slide10

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/sSlide11

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.Slide12

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?Slide13

Assignment

P. 156 #13-17, 21, 23