Circular Motion/Banking Forces

[calwyn]

1. A car of mass 1000kg can take a corner of 50m radius at 10ms-1.
a) Calculate the frictional force provided by the tyres that allows the car to take the corner.

The same car now takes a corner of the same radius but the road is banked inwards at 30 degrees.

b) What is the maximum speed the car can take the corner on the same tyres?

The Attempt at a Solution

For part a i got 2000N, which is fine. Just need to clear up on part b, coz i don't seem to be able to get it right :(

I used F=Nsin30
= 10000 x .5
= 5000N

V^2= Fr/m
= 5000x50/1000
v = 15.8 ms^-1

 

[anathema]

[/b]

Hello, great job on calculating the frictional force for part a! Now, for part b, you are on the right track but there are a few things to consider. First, the formula you used for calculating the maximum speed assumes that the force of friction is the only force acting on the car. However, in this scenario, there is also a component of the car's weight acting perpendicular to the banked road. This component can be calculated using the formula F = mgcosθ, where θ is the angle of the banked road (in this case, 30 degrees).

Once you have calculated this component, you can subtract it from the total frictional force calculated in part a to find the net force acting on the car. Then, you can use this net force in the formula you used (F = ma) to find the maximum speed the car can take the corner. Keep in mind that the maximum speed will be lower than the speed calculated in part a, since there is now an additional force acting on the car.

I hope this helps! Keep up the good work.

 

[Moetasim]

For part b, we need to take into account the normal force and the centripetal force acting on the car in the banked turn. This can be done by using the equations F=Ncosθ and F=mv^2/r, where θ is the angle of the banked road and v is the velocity of the car.

First, we can find the normal force (N) using the equation F=ma, where a is the centripetal acceleration. We know that the centripetal acceleration is given by a=v^2/r, so we can rewrite the equation as F=mv^2/r. Plugging in the values, we get:

F=1000 x (10)^2 / 50 = 2000N

This is the same frictional force calculated in part a.

To find the maximum speed the car can take the corner, we can set the centripetal force equal to the normal force multiplied by the cosine of the angle of the banked road (cosθ). This is because the normal force and the centripetal force are perpendicular to each other, so we can use the cosine function to find the component of the normal force that acts in the direction of the centripetal force. The equation would be:

mv^2/r = Ncosθ

We can rearrange this to solve for v:

v = √(Ncosθ x r/m)

Plugging in the values, we get:

v = √(2000 x cos30 x 50 / 1000) = 17.3 ms^-1

So, the maximum speed the car can take the corner on the same tyres is 17.3 ms^-1. This is higher than the speed in part a because the banked road provides an additional component of the normal force that acts in the direction of the centripetal force, reducing the amount of friction needed to keep the car moving in a circular path.