Car driving on curve problem

[squintyeyes]

A car moves on a circular, flat track at a speed of 15.0 m/s. If the car's mass is 1150 kg, and the radius of the turn is 55.0 m, what is the frictional force between the car's tires and the road?
_____________

If the coefficent of static friction between the tires and the road is 0.930, what is the maximum speed that the car can have and still safely negotiate the turn?
_____________


is this what you need to do for the first part?
What do i need to do about the second part?

m = mass
u = coefficent of friction
g = 9.8 m/s^2
a = centripetal acceleration= (v^2)/r
r = radius
v = speed
To find the answer, you set

ΣF = ΣF
umg = ma
ug = (v^2)/r

 

[pgardn]

A car moves on a circular, flat track at a speed of 15.0 m/s. If the car's mass is 1150 kg, and the radius of the turn is 55.0 m, what is the frictional force between the car's tires and the road?
_____________

If the coefficent of static friction between the tires and the road is 0.930, what is the maximum speed that the car can have and still safely negotiate the turn?
_____________


is this what you need to do for the first part?
What do i need to do about the second part?

m = mass
u = coefficent of friction
g = 9.8 m/s^2
a = centripetal acceleration= (v^2)/r
r = radius
v = speed
To find the answer, you set

ΣF = ΣF
umg = ma
ug = (v^2)/r

What force is supplying the centripetal force keeping the car in a circular path? And now what is the equation for calculating centripetal force?

 

[pgardn]

You have the equation for centripetal acceleration. So multiply that by m...
F=ma... you have what a is for a mass moving in a circle.

 

[squintyeyes]

so would part 1 equal 4,704.545 N? what about part 2

 

[pgardn]

so would part 1 equal 4,704.545 N? what about part 2

Ok that number means when you are going 15 m/s in a car of the mass given in a radius of 55m that the friction is supplying almost 5000 N of force.

So now if we know how "sticky" the surface between the tire and road is we can figure out the maximum force that friction can exert so that the car can continue in its circular path. The sticky number is mu... 0.9.. whatever.

You know mu, you can figure out what the normal force is... if you multiply these two numbers together what do you get?

 

[squintyeyes]

the maximum friction force? maybe?

 

[pgardn]

the maximum friction force? maybe?

Yeps...

set that equal to mv^2/r and solve for the v...

 

[squintyeyes]

thanks