UNIT 4
PHYS 2001
Example 01
Example 02
Example 03
Example 04
Example 05
Example 06
Example 07

 

 

Circular Motion - Example 03

 

 

3 (a) A banked circular highway curve is designed for traffic moving at 60 kmh-1. The radius of the curve is 200 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)

Projectiles

Fig 3

Information given:
vt = 60 kmh-1= (60 X 1000 / 3600) ms-1= 16.7 ms-1
vt = 40 kmh-1 = (40 X 1000 / 3600) ms-1= 11.1 ms-1
r = 200 m

(a) If the object moves in a circle it must have an umbalanced force (Fc) towards the centre of the circle called the centirpetal force.

Fc = mv2/r

There are 2 physical forces on the car
1. The weight of the car W = mg acting in the vertical plane and at an angle θ.
2. The Normal reaction of the road R acting at a right angle to the road.

R can be resolved into Rv and Rh

Rv = R Cos θ and Rh = R Sin θ

This road is designed so that when the vehicle velocity is 60 kmh-1 (16.7 ms-1) the centripetal force is generated ONLY by the normal reaction i.e. there is no friction between the road and the vehicle.

From the statemnet 'Assume the cars do not have negative lift' since the car is not moving vertically then by Newton's first law:

Rv = W
Rv = mg
Now Rv = R Cos θ

So R Cos θ= mg ---Eq1
Cos θ= mg/R ---Eq2

Also Rh = R Sin θ ---Eq3
So Sin θ = R/Rh---Eq4

Dividing Eq4 by Eq2 we get
Sin θ / Cos θ = (R/Rh) / (mg/R)

Tan θ = (Rh/mg)

but Rh is the unbalanced centripetal force Fc = mv2/r

So Tan θ = (mv2/r)/ (mg)

Tan θ = (v2)/ (gr)

θ = Tan-1 (v2)/ (gr)

By design θ = Tan-1 (16.72)/ [(g)(200)]

θ = 8.1°

Projectiles

Fig 3

 

Now for vehicles moveing at 40 kmh-1 (11.1 ms-1)

The centripetal force Fc = mv2/r becomes less and as the car tends to slip down the slope a frictional force F acting up the place opposes it.

F can be resolved into can be resolved into F v and F h

Fv = F Sin θ and Fh = F Cos θ

Recall F = μR so Fv = μR Sin θ and Fh = μR Cos θ

So if the car is not moving vertically then by Newton's first law:

Rv + Fv = W
Rv + μR Sin θ = mg

R Cos θ + μR Sin θ = mg

mg = R Cos θ + μR Sin θ ---Eq5

Horizontally Fc = mv2/r = Rh - Fh

mv2/r = Rh - μR Cos θ

mv2/r = R Sin θ - μR Cos θ ---Eq6

Dividing Eq6 by Eq5 we get:

mv2/rmg = (R Sin θ - μR Cos θ) / (R Cos θ + μR Sin θ)

v2/rg = R (Sin θ - μ Cos θ) / R (Cos θ + μ Sin θ)

v2/rg = (Sin θ - μ Cos θ) / (Cos θ + μ Sin θ)

(11.1 ms-1)2 / (-200m) (-9.81 ms-2) = (Sin θ - μ Cos θ) / ( Cos θ + μ Sin θ)

6.3 X 10-2 = (Sin θ - μ Cos θ) / ( Cos θ + μ Sin θ)

(6.3 X 10-2) ( Cos θ + μ Sin θ) = (Sin θ - μ Cos θ)

{(6.3 X 10-2) ( Cos θ)] + [(6.3 X 10-2) ( μ Sin θ)] = (Sin θ) - (μ Cos θ)

{(6.3 X 10-2) ( Cos θ)] + [(6.3 X 10-2) ( μ Sin θ)] + (μ Cos θ) = (Sin θ)

((6.3 X 10-2) (μ Sin θ) + (μ Cos θ) = (Sin θ) - (6.3 X 10-2) ( Cos θ)

μ [(6.3 X 10-2) Sinθ + Cos θ)] = (Sin θ) - (6.3 X 10-2) ( Cos θ)

μ = [(Sin θ) - (6.3 X 10-2) ( Cos θ)] / [(6.3 X 10-2) Sinθ + Cos θ)]

μ = [(Sin 8.1°) - (6.3 X 10-2) ( Cos 8.1°)] / [(6.3 X 10-2) Sin 8.1° + Cos 8.1°)]

μ = 7.86 X 10-2

 

 

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Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.