Centripetal Force through Turns

[RiskX]

Hi,

I face some difficulties trying to solve the following problem:

"You’re sitting in the passenger seat of the car, approaching
a tight turn with a 10.0-meter radius. You know that the coefficient of static
friction is 0.8 on this road (you use the coefficient of static
friction because the tires aren’t slipping on the road’s surface) and that the
car has a mass of about 1,000 kg. What’s the maximum speed the driver can
go and still keep you safe?"

I didn't understand why we use the coefficient of static fricftion if the car is already on a run... Is that beacuse the surface has changed from a stright plane to a curved one?

 

[stallionx]

Hi,

I face some difficulties trying to solve the following problem:

"You’re sitting in the passenger seat of the car, approaching
a tight turn with a 10.0-meter radius. You know that the coefficient of static
friction is 0.8 on this road (you use the coefficient of static
friction because the tires aren’t slipping on the road’s surface) and that the
car has a mass of about 1,000 kg. What’s the maximum speed the driver can
go and still keep you safe?"

I didn't understand why we use the coefficient of static fricftion if the car is already on a run... Is that beacuse the surface has changed from a stright plane to a curved one?

Because it is that friction force that keeps the car in circular motion.

 

[gonzacf]

I didn't understand why we use the coefficient of static fricftion if the car is already on a run... Is that beacuse the surface has changed from a stright plane to a curved one?

I am not sure but I guess you mean why use the static coeficient instead of the dynamic.

Note that the wheels are spining and in fact their surface is statically adhered to the road surface. Only if the car looses grip (for example going too fast in the turn) and starts to slide the coeficient changes.

 

[stallionx]

I am not sure but I guess you mean why use the static coeficient instead of the dynamic.

Note that the wheels are spining and in fact their surface is statically adhered to the road surface. Only if the car looses grip (for example going too fast in the turn) and starts to slide the coeficient changes.

The dynamic friction as the car rotates along the circle should equal the driving motor force ( in opposite direction ) for tangential acceleration to be zero.

So that uniform circular motion criteria are met.

 

[rwisz]

Hi,

Thank you for reaching out with your question. I can understand your confusion with using the coefficient of static friction in this scenario. Let me explain why it is necessary to consider it in this situation.

Firstly, it is important to understand that the coefficient of static friction is a measure of the maximum frictional force that can be exerted between two surfaces in contact without causing them to slip. In the case of a car turning, the tires are in contact with the road's surface and the friction between them is what allows the car to turn without slipping off the road.

Now, coming to your specific question, the car is approaching a tight turn with a 10.0-meter radius. This means that the car will have to undergo a change in its direction of motion, which requires a centripetal force to be exerted towards the center of the turn. In this case, the frictional force between the tires and the road's surface is what provides the necessary centripetal force to keep the car on the road and prevent it from sliding off.

The coefficient of static friction is used to calculate the maximum frictional force that can be exerted between the tires and the road's surface. This, in turn, helps us determine the maximum speed at which the car can safely turn without slipping off the road. If the car were to exceed this speed, the frictional force would not be sufficient to provide the necessary centripetal force, and the car would slide off the road.

So, to answer your question, we use the coefficient of static friction in this scenario because it helps us determine the maximum safe speed for the car to turn without slipping off the road. I hope this explanation helps to clarify your doubts. Let me know if you have any further questions.