Acceleration in rolling without slipping 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.
 A: 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.
A: 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
A: 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

