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Consider an object rolling without slipping down an inclined plane. The center of mass translational acceleration of this object is in general not equal to the acceleration of the object if it were sliding down the incline.

This may be a basic question, but why? In either case, aren't the net forces on the object the same, and in the same direction? So why would the acceleration of the center of mass change? To change acceleration, an additional or different force is needed. What is that force in rolling without slipping? To my knowledge, forces are the same in both cases - friction up the incline, normal force perpendicular to incline, gravity vertically downwards.

Jean Pierre
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3 Answers3

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The forces are not necessarily the same, no.

  • In a sliding scenario, the object experiences kinetic friction against the motion.
  • In a rolling scenario, the (round) object experiences static friction at the contact point.

These two forces are in their nature and in their application different.

Also note, that this comparison cannot be done for the same object. An object that can experience rolling cannot experience sliding (with non-zero friction) - it must be shaped differently for that to take place, meaning that we can compare the same amount of mass but not the exact same object.

Steeven
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  • In the case of rolling, is the friction still not $u_{s} \cdot mg$? Why isn't the acceleration the exact same, just changing out $u_{k}$ for $u_{s}$? – Jean Pierre Feb 02 '23 at 14:12
  • @JeanPierre This is Amontons' Law, $$f_k=\mu_kn,.$$ This law only applies to kinetic friction. It does not apply to static friction, no. You might have seen a similar-looking formula for the limit value of static friction: $$f_{s,\mathrm{max}}=\mu_s n,,$$ but note that this is only a formula for the maximum limit before sliding starts, not the formula for the actual static friction at any given moment. Static friction takes whatever value is necessary to prevent the contact point from sliding - and that value is lower than kinetic friction. In contrast, kinetic friction is constant. – Steeven Feb 02 '23 at 16:57
  • (@JeanPierre Also, as a side note, be aware that even in the kinetic friction formula (Amontons' Law) you shouldn't assume the normal force $n$ to be $mg$. It might be, but only when you are dealing with horizontal surfaces and with only Earth's gravity acting. In general it could also easily be something else. In your case you are mentioning an incline, so a non-horizontal surface - so I would not expect the calculation to become $\mu\cdot mg$ for kinetic friction.) – Steeven Feb 02 '23 at 17:00
  • I meant $mgcos{\theta}$ sorry, but I see what you're saying. Why does static friction take exactly the value necessary for $a_{cm} = \alpha \cdot r$? Thanks; this has been helpful. – Jean Pierre Feb 02 '23 at 19:40
  • @JeanPierre The geometric bond $$a=\alpha r$$ ties, from the perspective of the centre, the acceleration $a$ of an edge point to the angular acceleration $\alpha$. This bond is always true for a rigid object. For pure rolling the contact point is stationary and thus "moves with the ground", so to speak, which from the perspective of the centre means that $a=\alpha r$ relates the acceleration of the ground with the angular acceleration. From the ground's perspective - which is an inertial frame - this corresponds to the centre moving with this acceleration, and so: $$a_{cm}=\alpha r,.$$ – Steeven Feb 02 '23 at 21:29
  • Got it, thanks again. – Jean Pierre Feb 03 '23 at 00:46
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Assume that in the sliding scenario friction is zero and in the rolling scenario friction is sufficient to prevent sliding entirely.

The acceleration in the sliding scenario is $g. sin(\theta)$ by normal straight line motion and the potential energy is converted entirely into kinetic energy.

In the rolling scenario the sliding force $mg.sin(\theta)$ is entirely opposed by the frictional force so motion is not possible directly. However, the object will roll if its centre of mass is forward of the contact point, and forward motion will be a side-effect of the rolling. This means that the potential energy is converted into rotational energy and kinetic energy. Therefore the kinetic energy component will be less and the velocity downhill will be less than with the sliding case.

I believe the acceleration of a rolling object depends on its exact shape. There are some other posts that discuss this. For example: Acceleration of ball rolling down incline

rghome
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  • My question is not that. My question is, why isn't the frictional force just $mgu_{s}$ in the case of rolling without slipping? – Jean Pierre Feb 02 '23 at 14:36
  • The frictional force at the contact point is transferred to all parts of the object through internal tension and this counteracts the acceleration due to gravity. – rghome Feb 02 '23 at 15:39
  • So what is the frictional force? If if it counteracts, how does acceleration occur? – Jean Pierre Feb 02 '23 at 19:33
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The equations:

I) Sliding

$$ m\,a_S=F+m\,g\sin(\alpha)$$

II) Rolling

$$ \left(m+\frac{I}{r^2}\right)\,a_R=F+m\,g\sin(\alpha)$$

thus: the force is the same but the accelerated mass is different. In case of rolling you have additional "mass" $~(I/r^2)~$ to accelerate.

$$a_S > a_R$$

  • $~I~$ Body inertial
  • $~m~$ Body mass
  • $~r~$ Roll radius
  • $~F~$ External force
  • $~\alpha~$ Incline angle
Eli
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