Rolling dynamics of round objects down an incline

If a cylinder is rolling down an inclined plane with inclination $\theta=30^\circ$ (with coefficient of static friction between cylinder and the plane $\mu_s=0.8$) without slipping and it has mass $12kg$ and radius $0.5 m$ then the frictional force $f$ required is to be calculated . If you calculate it (I have done the calculations too but I don't have time to write all that and that's not of much importance to the question) it comes out to be $f = (mg\sin\theta)/3 = 20N$ . (Take $g = 10m/s^2$)

Here's a hideous FBD :

In this case the MAXIMUM static friction possible is $f_{max}=0.8*12*10*\cos30^\circ$ = $48\sqrt3 N$ . This is greater than the weight $mg$'s $\sin\theta$ component which is $mg\sin\theta = 60N$ .

So shouldn't the friction be equal to $60N$ ? I mean isn't that what happens when the maximum static friction is greater than that applied force, which it itself adjusts to the applied force and cancels it. I am sure this is what happens with the non-round objects.

I know if friction would be equal to $60N$ then the cylinder would stop translational motion and it won't roll but start slipping instead as the torque provided by the frictional force will provide more angular acceleration than the required value for rolling and then ultimately this will lead to a very complicated motion.

So why would the friction adjust itself to $20N$? Just because this is the exact force required to make the cylinder roll?

But the friction doesn't calculate how much it should adjust to make something roll . Why would the friction care if something is rolling or not? And if it does then how does it "know" how much to adjust? It can be $60N$ (that would mean no rolling though). So is the question flawed or am I missing something?

• Friction force can be greater than $mgsin(\theta)$ because $mgsin(\theta)$ doesn't cause friction force in this case. Apr 25, 2016 at 20:12
• @lucas Then what does ? And if the frictional force will be greater than the $mgsin\theta$ then wouldn't that make the cylinder move upward the plane ? Apr 25, 2016 at 20:16
• yes it would :) but it can be less. The point is: don't think of friction either as $\mu F_N$ nor as the down-slope-$mg$-component. It adjust, as you call it, this is quite right. Why it does... well this is your question, actually, right :)
– Ilja
Apr 25, 2016 at 20:28
• Friction force is a fixed force. It is fixed to the ground at each point. It doesn't move and thus it cannot make the cylinder move upward the plane. Apr 25, 2016 at 20:34
• @lucas Yes but if its greater than $mgsin\theta$ , then it would . Apr 25, 2016 at 20:37

First of all: your solution is correct, the friction is one third of what it would have been for a "non-round", as you call it, body.

How does friction know. I like such questions, but I don't see a convincing simple reasoning right now (but there probably is one), so I will use a little of computation. I hope it's different (less computating^^) than your solution. And I hope you can follow my wishy-washy treatment, it should convey the idea to someone who can compute it by himself...:

You probably accept, that friction wants to achieve $v=\omega r$, right? Otherwise there would be slipping. Now a special property of the system comes in: the moment of inertia of a cylinder has the factor $1/2$. This means, that the angular momentum is $L=\frac 12 pr$ (it would have been equal to $pr$ for a point) if $v=\omega r$.

If this is valid all the time, the derivatives have to be equal, too. So for the total force and the total torque you also get the ratio $\tau = \frac 12 Fr$. The torque (around the center) comes only from friction (times $r$ of course), whereas the net force is the downhill gravity component $\frac{mg}2$ minus the friction. So if you want them to have the ratio of 1:2 you have to adjust friction to one third of the downhill component. If it were different, then velocity and angular velocity would change not in the right proportion to keep equal. Which means slipping.

... as another viewpoint:
You can describe the combined translation and rotation around the center by just one rotation at each instant. It rotates around the boundary point (which moves, but at each instant this description is true). Now you can do analogous arguments with respect to this point, only with a moment of inertia of $\frac 32$

So the point is: $v=\omega r$ in this case is the equivalent of not moving in the non-rotating case. By the way you can at once compute the necessary friction if the body is not a cylinder but a sphere: then torque and force have to be at the ratio 2:5, you need a friction force of $\frac 27$ of the gravity component $\frac {mg}2$

• Thank you for the answer ! I got that friction wants $v=ωr$ so that there is no sliding motion . But the working behind how friction adjusts itself for it is not clear to me. Apr 25, 2016 at 21:05
• Well, it's the same as e.g. the tension in a rope that you pull. The rope adjusts its length a little (it's like a very stiff spring) so that its tension is equal to the force you pull it with. If it were not, it would move in the direction where the force is greater, and thereby reduce or increase its tension.
– Ilja
Apr 25, 2016 at 21:11
• Do you see the analogy? It's the same with friction (it comes from some tiny displacement of the material that causes internal stresses) - but it's harder to put into words, because it's not so nicely one-dimensional.
– Ilja
Apr 25, 2016 at 21:12

Friction acts to oppose the two surfaces in contact from sliding past each other. It will be no greater than necessary to do that.

• Okay that explains why it will be 20 N . But can you elaborate on how friction would actually "know" that it has to adjust to 20 N so that no sliding takes place ? I know its kind of a stupid question . Apr 25, 2016 at 20:24
• What you're asking is equivalent to asking why if I push a box on a level surface with a 20 N force to the East don't I get a 60 N force pushing it to the West and accelerating it in the opposite direction that's its being pushed. Apr 25, 2016 at 20:27
• But in that case its easy to understand how the friction will self adjust itself to 20 N . In my question it isn't obvious ! I know the friction = 20 N is necessary for rolling but how can friction "know" it ? Also, in the translational motion if the applied force is less than the frictional force then the object doesn't do any translational motion . But in the question , it does both translational and rotational . Apr 25, 2016 at 20:35
• @ Sagar Kaushik : It is the same with a spring : it reacts to the force you pull it with. If you pull with 20N it will extend until the internal tension increases to 20N pulling back against you. If you pull with 60N it extends until the internal force is 60N pulling back against you. If you pull with 200N that may be more than the internal forces can provide, so the spring becomes deformed. The spring does not 'know' how much force you are using; it simply reacts. Pulling against friction is like that. Up to a point the friction can oppose motion. Beyond that the contact is broken. Apr 26, 2016 at 16:25
• @sammygerbil You are right . Its just that in the above case , its harder for me to visualise how friction does this. In just translational motion its clear that the friction must have the same value as the applied force (Newton's 3rd law) if the limiting friction is greater than the applied force . But in my question , not so much . Its also clear and easy in your spring example. But anyway thanks a lot. Apr 29, 2016 at 14:00

The frictional force is given by $f = \frac 1 3 mg \sin \theta$ when the no slipping condition ($a=r \alpha$) is satisfied.

If the component of weight down the slope, $mg \sin \theta$, was equal and opposite to the frictional force then there would no net force on the cylinder and so the centre of mass of the cylinder would not be accelerating ($a=0$).
However there would be a torque about the centre of mass and so there would be an angular acceleration of the cylinder ($\alpha$).
Thus the no slipping condition, $a=r \alpha$, would not be satisfied.