APPLICATIONS OFCentripetal Force

And Circular Motion

A CAR TURNING ON ABANKED CURVE

Banking of curves is the tilting of curved roadways to increase centripetal force that will keep motor vehicles on the road rounding the curve.

A CAR TURNING ON ABANKED CURVE

Q:How many forces are

acting on the car (assumingno friction)?

Engineers have learned to “bank” curves so that cars can safely travel around the curve without relying on friction at all to supply the centripetal acceleration.

A car is going around a friction-free banked curve. The radius of the curve is r.

5

A: Two-- Gravity and Normal

FREE BODYFN

Fg (weight)

DIAGRAM

The Centripetal Force:

FNx

FNx

The Centripetal Force points in the direction towards

The center

7

FN x or FN sin is the centripetal force.

r

mvFF NC

2

sin

FN cos or FN y since the car does not accelerate in the vertical direction, this component balances the weight (mg) of the car.

mgFN cos

FORMULAS:

8

Represents the angle of elevation/the banked curve

mg

rmv

F

F

N

N /

cos

sin 2

From

tan v

gr

2

𝜃=𝑡𝑎𝑛−1 𝑣𝑟𝑔

2

#1 #2

PROBLEM 5: Banked Curves(without friction)

• A car is to make a turn with a radius of curvature of 60 m at a speed of 27 m/s. At what angle should the road be banked for the car to make the turn without relying on friction?

SOLUTION:Given:

V=27 m/s

R=60 m

Find: 51=

𝜃=𝑡𝑎𝑛−1 𝑣𝑟𝑔

2

A CAR TURNING ON ALEVEL ROAD

This situation is experienced when a car is turning a corner.

A CAR TURNING ON ALEVEL ROAD

The car exhibits UCM as it turns the corner. A centripetal force acts, which is supplied by the FRICTIONAL FORCE between the car’s tires and the road.

𝐹 𝑐=𝐹 𝑓

For the car to make the turn without skidding, a minimum

coefficient of static friction (µ) must be present b/w the tires of the car and the road.

A CAR TURNING ON ALEVEL ROAD

𝐹𝑓 =𝜇𝐹𝑁

FORMULAS:

𝐹𝑁=𝑊=𝑚𝑔where

#1And since

𝐹 𝑐=𝐹 𝑓

𝑚𝑣2

𝑟=𝜇𝐹𝑁#2

𝐹𝑓 =𝜇𝑚𝑔#1

FORMULAS:

𝑚𝑣2

𝑟=𝜇𝐹𝑁

𝑚𝑣2

𝑟=𝜇𝑚𝑔

#2

𝝁= 𝒗𝒓𝒈

𝟐

#3

PROBLEM 5: Turning on a Level Road

• A car is traveling at 9m/s in a circle that has a radius of 60 m. What must be the minimum value of µ for the car to make the turn without skidding?

SOLUTION:Given:

V=9 m/s

R=60 m

Find: µ 0.14=

𝝁= 𝒗𝒓𝒈

𝟐

The Daytona 500 is the major event of the NASCAR season held in Daytona, Florida. The turns in this nearly circular track have a maximum radius(at the top) of 316m and are banked steeply at 31 degrees. Suppose these turns were frictionless. At what speed would the cars have to travel around them?

BANK CURVES

SOLUTION:

sm

smm

rgv

/43

31tan)/80.9)(316(

tan

2

tan v

gr

2

PROBLEMS

#2

A motorcycle and its rider weighing 120 kg rounds a curve

of 80 m radius at 22 m/s.

5. What is the force of friction required for the cyclist to

go around the curve at this speed

(a)614 N (b) 726 N

(c) 885 N (d) 946 N

PROBLEMS

#3

A motorcycle and its rider weighing 120 kg rounds a curve

of 80 m radius at 22 m/s.

6. What is the frictional force needed to make a successful

turn if the coefficient of static friction is 0.65

(a) 622 N (b) 764 N

(c) 800 N (d) 948 N