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Note:- This is not the first time this question has been asked (far too many to link), but everytime it was asked as "why does friction cause a car to turn?" where rolling and rotation of the wheels got involved and made everything complicated. None of them seem to answer the more basic question that I have in mind.

I have been taught in physics class that static friction acts opposite to the direction of impending relative motion while kinetic friction acts opposite to actual relative motion, which seems to be the most general statement regarding friction direction. However lets consider a block (pt. mass) kept on a rough disk. The disk is spinning at a uniform constant angular velocity $\omega$ and the block is dropped onto the disk, initially at rest (the block not the disk). By even common sense we can see that the block will speed up until it has the velocity $\omega r$ where r is the distance of the block from the centre of the disk. In the steady state, the block spins along with the disk.

Q: Initially, why does the friction cause the block to go in a circle around the centre? (I think I understand what happens in steady state: the direction of impending motion is radially outward in the disk frame so friction acts radially inward.)

Qa): Is the statement "static friction acts opposite to the direction of impending relative motion" always correct?

Qb): What is happening in the case of a car turning on an unbanked road? Here the direction of ACTUAL relative motion is tangential and in the direction of car velocity, why should friction act radially instead of tangentially? Also is that force rolling friction?

PS: I am sorry for the word salad, will break into smaller qs if community says so.

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  • $\begingroup$ When the disc rotates and the block is at rest, why should there be a force in radial direction? I guess the block is hold at constant radius by a robe, is that correct? If so, please draw a picture and make the radius "large". Now ask yourself: If the block is at rest, in what direction does the disc underneath the block move? In radial direction? $\endgroup$ – Semoi Dec 14 '19 at 20:09
  • $\begingroup$ @Semoi No there is no rope. Only force acting is friction. $\endgroup$ – aditya_stack Dec 15 '19 at 6:12
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Q: Initially, why does the friction cause the block to go in a circle around the centre? I think I understand what happens in steady state: the direction of impending motion is radially outward in the disk frame so friction acts radially inward.

As you say, it opposes relative motion. The block is at rest and the disk is not. Friction creates a force that accelerates the block in the direction of rotation. As the block starts moving, the part of the disk underneath the block is pulled by the rigid forces within the disk to rotate (rather than go off in a straight line). Since the disk is accelerating, frictional forces drag the block to accelerate around the axis as well.

Qa): Is the statement "static friction acts opposite to the direction of impending relative motion" always correct?

As long as you properly identify the tendency, then yes.

Qb): What is happening in the case of a car turning on an unbanked road? Here the direction of ACTUAL relative motion is tangential

When we talk about the tire and the road, there is no relative motion. The contact patch of the tire remains still against the ground. There is no relative motion, so there is no kinetic friction, only static friction. Do not be confused by the motion of the car as a whole. It is the motion of the bottom of the tire that matters here.

Further, the wheel is an interesting device. At the ideal, a wheel has zero friction perpendicular to the axis (it rolls instead) and has infinite friction parallel to the axis. A real wheel will slip instead, but that's the intent.

...and in the direction of car velocity, why should friction act radially instead of tangentially?

Now you know, because that's what a wheel does. Unless it's hooked up to an engine or a brake system, the wheel responds to forces in the direction of the car by rolling. This is easy to do, so no forces arise.

But "sideways" forces on the wheel are created by friction at the contact patch. The wheel cannot roll in that direction, so the forces can be very large.

Also is that force rolling friction?

No. "rolling friction" is just one of the many sources of drag on the moving vehicle. There are many answers that discuss how it's different from static and kinetic friction.

Why is sliding friction more than rolling friction?

What is the cause of rolling friction? & why is it less than sliding friction?

Is acceleration of the disk necessary here?

Yes, because it's moving in a circle, not a straight line. Circular motion always requires (centripetal) acceleration.

so the centripetal force causing the wheel to turn, then, is static friction?

I'm not sure what you mean by "turn" here. The wheel can spin (allowing the car to roll forward or backward), and the steering wheel can cause the front wheels to pivot so that car turns in one direction or the other. Neither are caused by a centripetal force.

The wheel spins because the engine of the car has given the car forward velocity. Static friction on the road keeps the bottom of the wheel stationary against the road, so the spin of the wheel and the speed of the car are in sync. But this friction is not a centripetal force.

The front wheels turn sideways when a torque from the steering wheel pivots them. If the car is rolling forward, this causes a sideways force on the car and it creates a centripetal force that makes the car drive in a circle.

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  • $\begingroup$ Thank you for your clear answer but I still have a few doubts:- Q: "Since the disk is accelerating, frictional forces drag the block to accelerate around the axis as well." This is the only line I in Q I didn't fully understand. Is acceleration of the disk necessary here? $\endgroup$ – aditya_stack Dec 15 '19 at 5:57
  • $\begingroup$ Qa: "As long as you properly identify the tendency, then yes." meaning? $\endgroup$ – aditya_stack Dec 15 '19 at 6:03
  • $\begingroup$ Qb: "infinite friction parallel to the axis. A real wheel will slip instead, but that's the intent." why is there infinite friction parallel to the axis, and what does 'intent' mean? $\endgroup$ – aditya_stack Dec 15 '19 at 6:04
  • $\begingroup$ Finally, so the centripetal force causing the wheel to turn, then, is static friction? $\endgroup$ – aditya_stack Dec 15 '19 at 6:06

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