Consider a Disc of mass $M$ and radius $R$, I applied force $F$ tangentially on it. Now using $F=Ma$ , acceleration comes up to $$a=F/M$$

Now, let's use the torque equation: Here, the moment of inertia $I$ is $\frac12MR^2$ , and let $\alpha$ be the angular acceleration. Now, torque equals $FR$, so $$FR=\frac12MR^2\alpha$$ and, putting the rolling without slipping assumption $a=\alpha R$, we get $$a=2F/M$$ What gives rise to this discrepancy?

  • 4
    $\begingroup$ in rought terms, the first is for the acceleration of the center of mass, the second gives rotation about the center of mass, they are different things $\endgroup$
    – user65081
    Commented Jan 5, 2019 at 19:53
  • $\begingroup$ $a$ and $\alpha$ don’t even have the same dimensions. The former is an acceleration, the rate of change of velocity. The latter is an angular acceleration, the rate of change of angular velocity. $\endgroup$
    – G. Smith
    Commented Jan 5, 2019 at 20:09
  • $\begingroup$ I know the dimension thing. But I'm asking that, when we use F=ma , then we get acceleration as F/m. But, By using Torque equation , then using formula a=alpha*radius , acceleration's value changes by the shape of object, as it depends on moment of inertia. so, Why ??? $\endgroup$ Commented Jan 5, 2019 at 20:26
  • $\begingroup$ The $a$ in $F=ma$ is the acceleration of the center of mass (if $F$ is the total external force on the object). The $a$ in $a=\alpha r$ is the tangential component of the acceleration of a point on the object at distance $r$ from the axis of rotation. They are two different points and there is no reason to expect their acceleration to be the same. $\endgroup$
    – G. Smith
    Commented Jan 5, 2019 at 21:26
  • $\begingroup$ @G.Smith It is the relation for when things roll without slipping. $\endgroup$ Commented Jan 6, 2019 at 7:03

4 Answers 4


The other answers are missing the point (although I do recognize that some of those answers were posted before the question became more focused). By using the equation $a=\alpha R$ you are assuming rolling without slipping between the object and the surface the object is rolling across. What this means is that if you apply your own tangential force to the edge of the object, then to prevent slipping there must be, in general, a friction force acting on the object as well. So you now have two forces acting on your object, and the scenario is more complicated than it seemed at first (work done below).

The point is that if you use $a=\alpha R$, then you are limiting yourself to specific scenarios where this can happen. Right now you are working with conflicting assumptions. A disk rolling without slipping with only one tangential force acting on the edge of the body is an impossible scenario. As you have found, the disk would not be moving linearly fast enough to keep up with the spinning in order to have no slipping.

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 rolling 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.

So our net force is actually $$F_{net}=F+f=ma$$

And the net torque is $$\tau_{net}=R(F-f)=I\alpha=\frac aRI$$ For nice bodies like disks, spheres, etc. the moment of inertia takes the form $I=\gamma mR^2$, so we have $$F-f=\gamma ma$$

Combining the equations we get $$\gamma(F+f)=F-f$$ or $$f=\frac{1-\gamma}{1+\gamma}F$$

Now this might seem backwards from what you were trying to do. You were trying to determine the acceleration. The problem is that by using $a=\alpha R$ you are requiring the friction force to be a specific value, which we have found here. Then you can see that everything is now consistent:

For your disk, $\gamma=\frac12$, and so in order to have rolling without slipping the friction force must be $f=\frac13F$. With the forces of $F$ acting on top and $\frac13F$ acting on the bottom of the disk, we end up with a linear acceleration of $a=\frac{4F}{3m}$ which is consistent with either method you wanted to use. (In general, the acceleration, consistent with either method, is $a=\frac{2F}{(1+\gamma)m}$)

Bonus thoughts after working through this problem

1) For a ring $\gamma=1$, so you actually wouldn't need a friction force to have rolling without slipping if you did this with a ring

2) You wouldn't need friction with the disk if you applied your force a distance of $\frac12R$ vertically above the center of the disk.

3) Making point 2 more general, if we apply the force a fraction $\beta$ of the radius $R$ vertically above the center of the object, we end up with $$f=\frac{\beta-\gamma}{1+\gamma}F$$ which actually shows that friction could act in the opposite direction depending on where you apply the force.


For translations, it doesn't matter how the mass of a body is distributed, the acceleration will be $a=F/m$.

For rotations, the distribution of the mass is important. A ring whith large radius is harder to get into rotation than a small ring with the same mass. This is captured by moment of inertia.

  • $\begingroup$ Ya, u are right. BUT, I'm talking about same body. Consider a Disc of mass $M$ and radius $R$, I applied force $F$ tangentially on it. Now using $F=Ma$ , acceleration comes up to $F/M$. Now Lets use Torque equation. Here, $I$ (Moment of Inertia) is $(MR^2)/2 $ , and let $x$ be the angular acceleration. Now , Torque equals $F*R$, so equation will be $ F*R = (MR^2)/2 * x$ and, putting $a=x*R$, we get acceleration equals to $2F/M$. Why ??? Let me know. $\endgroup$ Commented Jan 5, 2019 at 21:23
  • $\begingroup$ a=xR is wrong because for a disc, R is not constant. You have to be careful when mixing rotation and translation. $\endgroup$
    – Jasper
    Commented Jan 5, 2019 at 21:36
  • $\begingroup$ Why R is not constant? At tangential point, it is always constant I guess??? $\endgroup$ Commented Jan 5, 2019 at 21:40
  • $\begingroup$ The mass (the thing opposing accelerations) is distributed over the whole disc. There is some mass at distance R, but also mass at smaller distances. $\endgroup$
    – Jasper
    Commented Jan 5, 2019 at 21:54
  • $\begingroup$ @Jasper The OP is assuming rolling without slipping, where the acceleration of the center of mass $a$ will be equal to the angular acceleration of the object multiplied by the radius of that object. $\endgroup$ Commented Jan 6, 2019 at 6:01

An object resists motion (acceleration) due to its inertia. When the change in velocity is linear, the inertia of the object depends simply on the amount of mass. When the change in velocity is rotational (angular velocity), like for a disc or wheel, its inertia depends not only on the amount of mass, but also on the distribution of the mass, with the mass farthest from the center of rotation typically contributing the most to its inertia. For example, the area moment of inertia of a disc is proportional to the fourth power of its radius.

A flywheel is a good example. It stores considerable kinetic energy due to its moment of inertia. Often times most of the mass is located at the perimeter of the wheel where the contribution to the moment of inertia is greatest.

Hope this helps.


The slightly simplified answer is that different points on the object can traverse different paths of different lengths in the same amount of time - their accelerations can be completely different depending on where they are in the object.

An explanation that I should hope clear things up a bit more is that, when dealing with classical mechanics problems, the versions of Newton's laws (that you imagine as being the most fundamental and universal) that you should hold in your mind should not be "the rate of change of momentum of a body..." or "the force exerted by object A", but rather versions phrased with a clearer focus on point-particles.

As an example, instead of

The acceleration vector of an object is proportional to the net-force vector on the object.
(a paraphrasing of Newton's $2^{nd}$ law)


The acceleration vector of a point-particle is proportional to the net-force vector on the object.

If you've never heard of the idea of a point-particle before, it may be a bit tough to wrap your head around at first, and its use may seem mundane. But, generally, it does help provide a better, more consistent understanding of (in particular) concepts like rotational motion and why we see such a difference between the equations for rotational motion and those for linear motion.

Most textbooks on mechanics beyond a high school level (at least that I know of) take the approach of introducing Newton's laws for these simpler point-particles and then prove more general results for rigid bodies.

Edit: An elaboration based on the concepts I mentioned - the problem is you can't really say F=ma for a rigid body. Okay fine, you sort of can, but the "version" of F=ma (if you want to call it that) for rigid bodies really just states that the net external force on a rigid body equals the body's mass times the acceleration of its center of mass.
(I call this a "version" of Newton's Second Law as really you can derive this from the more fundamental, unconditional (from a classical mechanics standpoint), and unambiguous, F=ma for point-particles).

That equation, of course, is completely useless here as the center may very well be fixed in place meaning that even if you just wanted to chuck this equation at your problem it wouldn't do you any good (the net external force will be 0 as long as you don't forget the force acting at the center itself to hold the disk in place) (also, there are problems where the center of mass of a body does move and its motion can provide useful info about rotation but those are more complicated and, in a sense, special cases).

Suppose one were to, instead, examine a small segment (as a point particle approximation) on, say the rim of a disk rotating, where (e.g.) a frictional force is being applied, it would not be so simple for me to conclude that $friction=mass_{segment}*a_{segment}$ (we want non-zero mass which is why I mention a segment). In fact, this would be wrong!-as even for point particles, while Newton's $2^{nd}$ law does say F=ma, the F here is specifically referring to the net force. And if you think about it, well, aren't there a ton of other point-particles (tiny segments, if you like) in the disk that might each be exerting their own force on the rim segment? Indeed, there are. But how on Earth are you supposed to calculate all of those?! There are also many other complications that may arise, like if the friction is being applied at a fixed point while the body is moving, meaning that the friction acts on different segments at different times.

A path around all of these problems is given by equations like $\tau=I\alpha$ and the equations for kinetic energy conservation for rotating bodies which all deal with gross, knowable quantities like net external torque and angular velocity- but, we come to all of these equations via a treatment of the problem of rigid body rotations using the idea of torques for point-particles. Note, that no physics foreign to Newton's laws is introduced in rotational mechanics (although, some simplifying assumptions are sometimes used- for instance, derivations of $\tau=I\alpha$ generally just assume only central forces to be present in the body).

  • $\begingroup$ Can you explain me this, Mathematically. Consider a Disc of mass M and radius R, I applied force F tangentially on it. Now using F=Ma , acceleration comes up to F/M. Now Lets use Torque equation. Here, I (Moment of Inertia) is (MR2)/2 , and let x be the angular acceleration. Now , Torque equals F∗R, so equation will be F∗R=(MR2)/2∗x and, putting a=x∗R, we get acceleration equals to 2F/M. Why ??? Let me know. $\endgroup$ Commented Jan 5, 2019 at 21:49
  • $\begingroup$ Can you include this (which seems to be the real question here) in the Question with proper formatting? $\endgroup$
    – Jasper
    Commented Jan 5, 2019 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.