Why is the direction of friction different in case of rolling on plane surface and on an inclined plane?

I was studying the rolling of spherical objects on plane surfaces and inclined planes. I had doubts about the direction of friction in both cases.

Case 1- In the first case i.e. rolling on the plane surface, the friction acts in the direction of motion of the object. I tried searching why does this occurs and found an answer. It explained that

Let's assume that we have our rolling object (let's say it's not necessarily a disk) initially at rest, and then we apply a tangential force F to the top of it to start it rolling. However, if we want to roll without slipping, then we need an additional friction force f acting at the bottom of the object. This friction force will act in the same direction as our applied force because if there was no friction the object would spin too fast, so friction needs to decrease the net torque.

Case 2: Now in the case of rolling the friction acts in the backward direction and the reasoning used for this was that friction acts in a backward direction to prevent slipping...

But I couldn't get the reason used to explain this. So it would be helpful if someone could explain the difference in these two situations and answer that why is the direction of friction different in both the cases?

• I just want to know that why the direction of friction is different in both cases and what exactly is the difference between these two situations. Oct 24 '21 at 6:43

in order for rolling without slipping to occur, the linear acceleration has to match the angular acceleration by $$a=\alpha r$$

For the first case, if there was no friction, the linear acceleration is given by $$a=\frac Fm$$ and the angular acceleration is given by $$\alpha=\frac{Fr}{I}$$ If we have a nice object such that $$I=\gamma mr^2$$, where $$\gamma$$ is some dimensionless constant, then the angular acceleration is $$\alpha=\frac{F}{\gamma mr}\to\alpha r=\frac{F}{\gamma m}$$

Since $$\gamma<1$$ for something like a sphere, we see that $$a$$ is too small and/or $$\alpha$$ would be too large, so rolling with slipping would occur without friction. Therefore, in order for friction to allow for rolling without slipping, we need the friction force to raise the net force while lowering the net torque. This is achieved on the bottom of the object by friction acting in the same direction as the force $$F$$.$$^*$$

In the inclined plane example, the force acting along the incline is the component of weight along the incline, but we take that to act at the center of the object, therefore, without friction there would be no rotation at all and sliding would certainly occur. Therefore, friction needs to act in order to increase the net torque but decrease the net force: hence it acts up the incline.

In either case the same concepts are applied: rolling without slipping requires $$a=\alpha r$$.

$$^*$$Note that for a circular hoop $$\gamma=1$$, and so there actually would not be any friction force in this case; rolling without slipping would occur automatically even on a frictionless surface.

• Since γ<1 for something like a sphere, we see that a is too small and/or α would be too large, so rolling with slipping would occur without friction. I actually didn't get this line...If γ<1 then wouldn't α be smaller and a be greater...as let us assume that F/ (γm)= p...so α=p/r and a =p...so we have α divided by r which will make it smaller... Oct 29 '21 at 15:09
• @SakshamSalaria Dividing by a number smaller than $1$ increases the value. So $\alpha r>F/m$ Oct 29 '21 at 16:07
• So have we assumed this??? Oct 30 '21 at 13:21
• @SakshamSalaria for "nice objects" where $I=\gamma mr^2$, $\gamma$ cannot be larger than $1$. $\gamma>1$ implies that mass exists outside of the body, which doesn't make sense. Oct 30 '21 at 13:47
• Could you please tell me what exactly γ is? Thanks for helping me out btw... Oct 30 '21 at 17:31

So it would be helpful if someone could explain the difference in these two situations and answer that why is the direction of friction different in both the cass?

The difference is where the "driving" force is being applied. In the first case the force is being applied at the top of the wheel, and in the second case the force is being applied at the center of the wheel.

As described by @BioPhysicist, for rolling without slipping we have the condition that $$a=\alpha r$$. Assume that we have a known driving force $$F$$ applied in the positive direction a distance of $$h$$ from the center of the wheel where $$-r and positive $$h$$ is above the center. There is also an unknown friction force $$f$$ applied in the positive direction at the road. Then by Newton's 2nd law (both rotational and linear) we have: $$F + f = ma$$ $$F h - f r = I \alpha = Ia/r$$ This gives two equations in two unknowns, $$f$$ and $$a$$. We can easily solve those to obtain: $$f=F\frac{hmr-I}{mr^2+I}$$ $$a=F\frac{hr+r^2}{mr^2+I}$$ or for "nice" objects such that $$I=\gamma m r^2$$ we get $$f=F\frac{h-r\gamma}{r+r\gamma}$$ $$a=\frac{F}{m} \frac{h+r}{r+r\gamma}$$

We see that if $$r\gamma < h$$ then $$f$$ is in the same direction as $$F$$. For a disk $$\gamma=1/4$$ so the first example with $$h=r$$ gives $$r/4 < r$$ so they are in the same direction while the second example with $$h=0$$ gives $$r/4 \nless 0$$ so they are in opposite directions.

• Oct 29 '21 at 18:08
• @BioPhysicist yes, I was looking for that post, but couldn't find it!
– Dale
Oct 29 '21 at 19:36

An un-powered circular object rolling on a horizontal surface experiences only rolling friction due to the deformation of the two surfaces in contact. It's like the object is trying to roll uphill. If a wheel is powered by an axle a static friction force from the road acts in the forward direction. If the breaks are locked on a moving wheel, a kinetic friction force will act in the backward direction. An object rolling freely down an incline will need a backward static friction force to keep the angular acceleration in sync with the linear acceleration. On a friction-less surface (without deformation) there will be no link between linear and angular acceleration.