Centripetal force .

[siddharth5129]

Main Question or Discussion Point

My textbook says that the requisite centripetal force for the circular motion of a vehicle is provided by friction. I dont get that ............. i mean , shouldn't friction oppose relative motion ..........so , seeing as the direction of motion of the car is along the tangent to the curve , shouldn't friction act along the tangent to the curve in the opposite direction...what gives?

 

[ZapperZ]

If there's no friction, the vehicle cannever make that turn. Or if the frictional force is LESS than the needed centripetal force, the vehicle will slide. Guess which direction it will slide?

In any case, this is all hand-waving. All you need to do is draw a free-body diagram and account for all the forces involved. If you follow what you had in mind, you'll see something missing, i.e. where is the source of the centripetal force that causes the vehicle to move in a circular path?

Zz.

 

[S_Happens]

In an instantaneous sense you can consider the car to be moving tangentially to the curve, but that is exactly the type of thinking that got you stuck.

Consider what the car does through the entire turn. What does the car have to do to turn (not meaning "turn the wheels," but rather is there a force need be applied)?

Edit- wow, beaten to a response at 4 AM...

 

[Doc Al]

My textbook says that the requisite centripetal force for the circular motion of a vehicle is provided by friction. I dont get that ............. i mean , shouldn't friction oppose relative motion ..........so , seeing as the direction of motion of the car is along the tangent to the curve , shouldn't friction act along the tangent to the curve in the opposite direction...what gives?

Careful. Just because the car is moving to the right (say) doesn't mean friction acts to the left, since the tires are rolling. It's not the same as if you were to drag a block of wood to the right--in which case friction would act to the left.

Friction opposes slipping between surfaces. The contact surfaces are the tire and the ground. Realize that the tire (presumably) is rolling without slipping--thus the friction involved is static friction. As ZapperZ says, you must analyze the forces acting on the car and its acceleration to understand the direction of the friction force.

 

[Caesar_Rahil]

shouldn't friction oppose relative motion

well in a sense, i think it is opposing relative motion,
for that u hav to analyse the point where friction is acting..
if u can see it, car bending on a right turn actually have a tendency of keeping to the left due to inertia,
friction opposes this.

also, there is a difference between rolling, static and kinetic friction...
while turning, static friction is acting as the point of application of force is always at rest
i.e it does not RUB or SLIP on the ground.
kinetic friction acts when slipping occurs.
Rolling friction , is suppose is irrelevant as we can assume road is hard

 

[siddharth5129]

well in a sense, i think it is opposing relative motion,
for that u hav to analyse the point where friction is acting..
if u can see it, car bending on a right turn actually have a tendency of keeping to the left due to inertia,
friction opposes this.

also, there is a difference between rolling, static and kinetic friction...
while turning, static friction is acting as the point of application of force is always at rest
i.e it does not RUB or SLIP on the ground.
kinetic friction acts when slipping occurs.
Rolling friction , is suppose is irrelevant as we can assume road is hard

I can see your point about the car having a tendency of keeping to the left due to inertia ......but i don't get why the point of application of force is always at rest ..... , isn't it moving along with the car ....so why does static friction act?

 

[Bob S]

The frictional (centripital) force opposes the centrifugal outward force. Neither of these is in the instantaneous direction of motion dx, so their vector product with dx is zero, and so no work (F*dx) is done and the vehicle velocity ideally does not change.