Acceleration question (circular motion)

[SmallPub]

Homework Statement

will the acceleration of a car be the same when the car travels around a sharp curve at a constant 60 km/h as when it travels around a gentle curve at the same speed? Explain.

Homework Equations

None

The Attempt at a Solution

No the acceleration will not be the same because when you are turning the car sharply so you are changing your direction right away. If you are turning at a gentle curve at the same speed you are changing your direction not as fast as a sharp turn so the acceleration would be different. Am I correct?

 

[BOYLANATOR]

You are correct. There is an equation you could use to confirm your answer. If you assume that while the car travels around the bend it is briefly in circular motion, what formula could you use to describe the acceleration?

 

[SmallPub]

You are correct. There is an equation you could use to confirm your answer. If you assume that while the car travels around the bend it is briefly in circular motion, what formula could you use to describe the acceleration?

I was thinking ac = v^2/r , ac= 4∏^2r/T^2

But I am not sure

 

[BOYLANATOR]

The first one looks good. Can you see what would happen to the variables in that equation if the corner became sharper?

 

[SmallPub]

The first one looks good. Can you see what would happen to the variables in that equation if the corner became sharper?

Would the radius become smaller making the acceleration larger? Correct me if I'm wrong

 

[BOYLANATOR]

Yes. For a bonus, why is the car more likely to slide off the road and crash on the sharper bend?

 

[SmallPub]

Yes. For a bonus, why is the car more likely to slide off the road and crash on the sharper bend?

There would be less friction to keep the car on the road, making the car crash.

 

[BOYLANATOR]

Right idea, but there's no real reason for the friction to decrease, it's more likely to remain constant...

 

[SmallPub]

Right idea, but there's no real reason for the friction to decrease, it's more likely to remain constant...

Ok got it! Thanks for the help!