Video Transcript
A car with mass 360 kilograms travels at constant speed along a circular path around a flat roundabout. The radius of the roundabout is 12 meters. The car takes a time of 28 seconds to completely travel around the roundabout. What is the friction force between the wheels of the car and the surface of the road? Give your answer to the nearest newton. What is the coefficient of static friction for the wheels of the car on the surface of the road? Use a value of 9.8 meters per second squared for the acceleration due to gravity. Give your answer to three decimal places.
We’ll look at the parts of our question in order, starting with this part where we want to solve for the friction force between the wheels of the car and the surface of the road. Let’s say that this is our roundabout with a radius we’ll call 𝑟 and a car with a mass we’ll call 𝑚 driving around on it. As the car drives along, it experiences a force towards the center of the roundabout. We call this the centripetal force on the car. In equation form, the centripetal force acting on an object with mass 𝑚 moving in a circle of radius 𝑟 is equal to 𝑚 times 𝑟 times the angular speed 𝜔 of that object squared.
An important note about centripetal forces is that there are always physical causes for them. A centripetal force on an object is just the force that object experiences that pushes it towards the center of a circular arc. For our car, the reason it experiences such a force is due to friction. Specifically, it’s the friction between the tires of the car and the surface of the road. That friction force on the car, we’ll call it 𝐹 sub f, is center seeking. That is, the frictional force is the force that pushes the car towards the center of the roundabout. Without the friction force acting on the car, it couldn’t follow this circular path.
It’s that friction force we want to solve for. And by our equation for centripetal force, we know it equals the car’s mass times the radius of the roundabout times the angular speed of the car as it travels squared. Simplifying our expression a bit, we know the mass of the car, and we’re given the roundabout radius. However, we don’t yet know the car’s angular speed as it travels. We can recall that, in general, angular speed 𝜔 equals a change in angular position 𝛥𝜃 divided by a change in time 𝛥𝑡.
Now, let’s imagine we follow our car as it makes one complete revolution around the roundabout. In that case, we would say that the change in angular position of the car is two 𝜋 radians. And then, in our problem statement, we’re told that it takes the car 28 seconds to make one complete revolution. So then, two 𝜋 radians divided by 28 seconds is equal to this car’s angular speed 𝜔.
Simplifying this fraction a bit, if we divide numerator and denominator by two, we get 𝜋 radians in 14 seconds. This is the value for 𝜔 that we can use to substitute into our equation for the friction force. The mass of our car is 360 kilograms. The radius of the roundabout is 12 meters. And 𝜔 is 𝜋 radians over 14 seconds. In terms of the units here, radians are dimensionless. So, the overall units we have are kilograms times meters per second squared. These are equal then to units of newtons. When we calculate this expression and round the result to the nearest newton, we get 218 newtons. This is the friction force between the wheels of the car and the surface of the road.
Let’s now look at part two of our question: What is the coefficient of static friction for the wheels of the car on the surface of the road? Use a value of 9.8 meters per second squared for the acceleration due to gravity. Give your answer to three decimal places.
Considering a side-on view of the car as it travels, we know that there’s a friction force exerted between the surface of the road and the car’s tires. Even though the car’s tires are rotating, this is a static friction force so long as the car isn’t skidding or sliding across the road surface.
Assuming that’s the case, the magnitude of the friction force depends on what’s called the coefficient of static friction between the wheels of the car and the road surface. In general, the frictional force 𝐹 sub f on some object equals a coefficient of friction represented by the Greek letter 𝜇 multiplied by what’s sometimes called the normal force and sometimes called the reaction force on that object. For our car on this flat roundabout, that normal or reaction force acts straight upwards, and it is equal and opposite to the weight force that acts on the car, the mass of the car times the acceleration due to gravity.
For our scenario then, we can write that the frictional force on the car equals the coefficient of static friction, that’s why there’s this s subscript on 𝜇 here, multiplied by the normal force on the car, which is equal in magnitude to the mass of the car times 𝑔. Since it’s the coefficient of static friction that we want to solve for here, we can rearrange this equation by dividing both sides by 𝑚 times 𝑔 to do that. On the right-hand side, the factors of 𝑚 and 𝑔 cancel from numerator and denominator. The coefficient of static friction equals the friction force on the car divided by its mass times the acceleration due to gravity.
We’ve either solved for or been given all of these values. We found the friction force 𝐹 sub f to be 218 newtons. The mass of the car is 360 kilograms. And 𝑔, we’re told, is 9.8 meters per second squared. Rounding this value to three decimal places, we get a result of 0.062. Notice that there are no units in this answer. This is because a newton, the unit in our numerator, is a kilogram-meter per second squared. That matches perfectly with the units in our denominator, so all units cancel out of our expression. Indeed, it’s always the case that when solving for a coefficient of friction, the result is unitless. So, to three decimal places, the coefficient of static friction for the wheels of the car on the surface of the road is 0.062.
