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There is a ball rolling down an incline, with no slipping. If we consider the point of contact between the ball and the inclined plane to be the pivot point (for our torque calculations), then I have noticed that friction is no longer in play, and gravity (and normal force) is the reason for the torque. However, if we consider the center of mass of the ball to be the pivot point, friction is the only factor, since both the normal force and gravity effect the center of mass (so the net torque caused by them is 0, since the pivot is the center of mass). However, both of these produce different torques.. for different reasons. So my question is,

Is gravity the reason a ball rolls, or is friction? And why?

[Note: This is not homework, it is a genuine question I had while looking at some physics problems.]

user3904840
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    Thought the torques are different,so are the corresponding moment of Inertia.Hence the physical observable,the angular acceleration remains the same.As for the second question,the ball falls down due to gravity and rolls due to friction as seen from CM frame.In other frames,the other forces provide the torque,e.g. in the point of contact frame,the "roll" force is gravity. – Sandesh Kalantre Dec 01 '14 at 03:09
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    The roll motion itself is frame-dependent,and so is it's cause. – Sandesh Kalantre Dec 01 '14 at 03:11
  • @SandeshKalantre then why, in the first case, is the torque caused by gravity? Shouldn't gravity have an effect on the rolling then...? – user3904840 Dec 01 '14 at 03:49
  • @SandeshKalantre you might write that as an aswer, it is the correct response. –  Dec 01 '14 at 06:45
  • @user3904840 the comment by Sandesh is correct, the cause of the torque depends on your system of coordinates. But that is a different question that asking what "physically" causes the rotation. The torque is mathematical tool to calculate the rotation, the actual causes are the interplay between friction and gravity, if one of the two were absent, there would be no rotation –  Dec 01 '14 at 06:48
  • @Wolphramjonny please explain how the torque will be 0 if there is no friction & one calculates the torque about the point of contact. – Shashaank Jan 10 '17 at 11:33

2 Answers2

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In these cases it always helps to draw a diagram:

enter image description here

The green vectors represent the force of gravity $w=mg$ (dashed) and its components along the inclined plane and perpendicular to it. The red forces are the normal force of the plane on the ball $n$, the force of friction $F$, and their vector sum (dashed).

Now the sphere rotates about the contact point - that is the point that doesn't move. In that frame of reference, noting that the red vectors all pass through the center of rotation we compute the torque as the force of gravity $w$ times the perpendicular distance to the pivot point $d= r\sin\theta$, i.e. $$\Gamma = w\cdot r \sin\theta$$ and we consider the moment of inertia of the ball about this pivot to be $$I = \frac25 mr^2 + mr^2=\frac75 mr^2$$ (by the parallel axes theorem).

As you pointed out, by considering the motion about the contact point, the value of $F$ doesn't seem to come into play. But remember that the center of mass of the sphere must accelerate as though all forces are acting on it; after canceling out the normal forces, that leaves us with $mg\sin\theta$ down the slope, and $F$ going the other way. The difference between these two forces gives rise to the acceleration of the sphere's c.o.m. so we can compute $F$ from

$$mg \sin\theta - F = m a$$

To compute $a$, we first need the angular acceleration $\dot\omega$which is found from

$$\dot \omega = \frac{\Gamma}{I} = \frac{mgr\sin\theta}{\frac75 m r^2} = \frac{5g\sin\theta}{7r}$$

The linear acceleration $a$ is of course the angular acceleration multiplied by the radius of the sphere, so

$$a = \frac57 g\sin\theta$$

From which it follows that

$$F = \frac{2}{7} m g \sin \theta$$

And if we know that, we can now compute the angular acceleration of the sphere about its center. The torque seen in the frame of reference of the sphere is

$$\Gamma' = Fr = \frac{2}{7} m g r \sin\theta$$

Now we use the moment of inertia of the sphere about its center in order to compute the angular acceleration, and find

$$\dot \omega = \frac{\Gamma'}{\frac25 mr^2} \\ = \frac{\frac{2}{7} m g r \sin\theta}{\frac{2}{5} m r^2}\\ =\frac{5 g \sin\theta}{7 r}$$

which is the same result as before.

So there is no contradiction. The forces of friction and gravity work together to cause the rotation - the difference in apparent torque comes about from the fact that you are working in different (and non-inertial) frames of reference, but if you do the calculation carefully you get the same answer.

Bill N
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Floris
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  • If there were no friction, then obviously the ball would not roll; yet, when you perform the computation isn't there a discrepancy between the frame of the COM and the frame of the point of contact? From the frame of the COM we get that the torque is zero (as it should be) since there is no frictional force and the torque from gravity is zero. However, from the frame of the point of contact there is only torque due to gravity which is non-zero... – 1110101001 Jan 07 '16 at 07:37
  • When there is no friction, the force of gravity acts on the center of mass so there is no torque. Torque appears due to the friction force only - it is the only force that does not pass through the center of mass and therefore the only one that can provide torque - regardless of the frame of reference. No friction, no torque. – Floris Jan 07 '16 at 11:01
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    If you choose the rotating point as the point of contact, though, all torque is due to gravity: $\vec{w} \times \vec{r}$. Friction doesn't seem to play a part when you choose the torque point as the point of contact. – 1110101001 Jan 07 '16 at 23:29
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    Torque is due to either a pair of equal and opposite forces, or a single force offset w.r.t. the center of mass. You can't just make torque appear by choosing an axis and ignoring that. – Floris Jan 08 '16 at 02:21
  • But what goes wrong when you make the computation $$\dot \omega = \frac{\Gamma}{I} = \frac{mgr\sin\theta}{\frac75 m r^2} = \frac{5g\sin\theta}{7r}$$

    Doesn't the above equation hold in the case without friction as well?

     – 1110101001 Jan 08 '16 at 04:30
  • No - the first part of the expression is correct, but the torque is not equal to $mgr\sin\theta$ in the frictionless case. As I have tried to express multiple times. – Floris Jan 08 '16 at 04:36
  • Pardon, but I still don't see why the torque wouldn't be $m g r \sin \theta$. I get that gravity acts through the center of mass, but that would only matter if the torque is taken about the center of mass, right? In an equilibrium situation it should not matter where the torque is taken about, and no matter the point chosen the net torque is the sum of F cross r across all external forces acting on the body. In this case, gravity exerts a perpendicular component of $mg \sin \theta$ at a distance of $r$. So why would the torque due to gravity be zero when taken about the point of contact? – 1110101001 Jan 08 '16 at 06:45
  • I have some doubts and it will be great if you can answer them. What I am confused is that the motion will be purely rotational about point of contact and translation+ rotation al about com. But wouldn't the point of contact already posses the acceleration and velocity of com due to the property of com so how come it is inertial.... – Scáthach Sep 04 '18 at 15:09
  • If the point of contact is stationarty then that would mean the point doesn't move which would mean the disc as a whole won't move as the point is stationary / fixed to the ground but clearly the point or the vertex changes as the disc move.. Isn't this contradictory too – Scáthach Sep 04 '18 at 15:12
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    @harambe the point of contact is “instantaneously” stationary- because it is in contact with the plane and there is no slip. However it has acceleration - so its velocity is zero for just an instant. Like a ball bouncing - there is a moment it is not moving, but right before and right after that moment it does move. – Floris Sep 08 '18 at 15:01
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If we consider the point of contact between the ball and the inclined plane to be the pivot point (for our torque calculations), then I have noticed that friction is no longer in play, and gravity (and normal force) is the reason for the torque.

You have two choices for that point. Do you mean the point at rest on the ramp (only touching for an instant) or the moving point that is always touchting?

For the first choice, gravity does indeed provide a torque, but that torque is not going completely into rotation. Some of it is going into changing the angular acceleration of the ball that is due to its motion. That momentum is frame dependent.

For the second choice, since the ball is accelerating, the contact point is at rest only in a non-inertial frame. As such, there are fictitious forces that appear. You need to add a pseudo-force that operates on the ball's center of mass in the opposite direction of the acceleration. When you add in that pseudo force, the torque matches that when calculated around the center of mass.

BowlOfRed
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