Answer:
The coefficient of static friction between the tires and the road is 1.987
Explanation:
Given:
Radius of the track, r = 516 m
Tangential Acceleration [tex]a_r[/tex]= 3.89 m/s^2
Speed,v = 32.8 m/s
To Find:
The coefficient of static friction between the tires and the road = ?
Solution:
The radial Acceleration is given by,
[tex]a_{R = \frac{v^2}{r}[/tex]
[tex]a_{R = \frac{(32.8)^2}{516}[/tex]
[tex]a_{R = \frac{(1075.84)}{516}[/tex]
[tex]a_{R = 2.085 m/s^2[/tex]
Now the total acceleration is
[tex]\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2[/tex]
=>[tex]= \sqrt{ (a_r)^2+(a_R)^2}[/tex]
=>[tex] \sqrt{ (3.89 )^2+( 2.085)^2}[/tex]
=>[tex] \sqrt{ (15.1321)+(4.347)^2}[/tex]
=>[tex]19.4791 m/s^2[/tex]
The frictional force on the car will be f = ma------------(1)
And the force due to gravity is W = mg--------------------(2)
Now the coefficient of static friction is
[tex]\mu =\frac{f}{W}[/tex]
From (1) and (2)
[tex]\mu =\frac{ma}{mg}[/tex]
[tex]\mu =\frac{a}{g}[/tex]
Substituting the values, we get
[tex]\mu =\frac{19.4791}{9.8}[/tex]
[tex]\mu =1.987[/tex]
