Mechanical Engineering Principles

MOTION IN A CIRCLE 183

h=vertical distance of the centre of gravity

of the car from the ground,

L=distance between the centre of the tyres,

r=radius of the turning circle, and

μ=coefficient of friction.

o

mg

h

L

r CG

R 1 R 2

F 1 F 2

CF

Figure 16.2

Problem 1. Determine expressions for the

frictional forcesF 1 andF 2 of Figure 16.2.

Hence determine the thrust on each tyre.

Resolving forces horizontally gives:

F 1 +F 2 =CF=

mv^2

r

( 16. 2 )

Resolving forces vertically gives:

R 1 +R 2 =mg ( 16. 3 )

Taking moments about the ‘outer’ wheel gives:

CF×h+R 1 ×L=mg

L

2

i.e.

mv^2

r

h+R 1 L=mg

L

2

or R 1 L=mg

L

2

−

mv^2

r

h

Hence, R 1 L=m

(

gL

2

−

v^2 h

r

)

from which, R 1 =

m

L

(

gL

2

−

v^2 h

r

)

( 16. 4 )

Also, F 1 =μR 1 andF 2 =μR 2 ( 16. 5 )

Substituting equation (16.4) into equation (16.3)

gives:

m

L

(

gL

2

−

v^2 h

r

)

+R 2 =mg

Therefore,

R 2 =mg−

m

L

(

gL

2

−

v^2 h

r

)

=mg−

mg

2

+

m

L

v^2 h

r

i.e. R 2 =

m

L

(

gL

2

+

v^2 h

r

)

( 16. 6 )

From equations (16.4) to (16.6):

F 1 =μ

m

L

(

gL

2

−

v^2 h

r

)

( 16. 7 )

and F 2 =μ

m

L

(

gL

2

+

v^2 h

r

)

( 16. 8 )

To calculate the thrust on each tyre:

From Pythagoras’ theorem,

T 1 =

√

F 12 +R^21 =

√

μ^2 R^21 +R^21

i.e. T 1 =R 1 ×

√

1 +μ^2 (see Figure 16.3(a))

Letα 1 =angle of thrust,

i.e. α 1 =tan−^1

(

F 1

R 1

)

=tan−^1 μ

From Figure 16.3(b), T 2 =R 2 ×

√

1 +μ^2

α 2 =tan−^1

(

F 2

R 2

)

=tan−^1 μ