# Why is Newton's second law seemingly not applicable to a ball rolling down incline plane?

A homework problem asked us to find the acceleration of a ball (pure) rolling down an incline plane without friction. I thought it was simply $$a=g \ \sin(\alpha)$$ where $$a$$ is the acceleration of the CM and $$\alpha$$ is the angle of the incline to the horizontal.

The solution says, however, that the acceleration (like in this related thread: Acceleration of ball rolling down incline) is $$(3/5) \ g \ \sin( \alpha)$$.

Since we are only interested in the acceleration of the Center of Mass and the only force is the parallel component of gravity (since there is no friction and the normal force compensates the perpendicular component), shouldn't Newton's second law gives us the simple $$g \ \sin(\alpha)$$ answer?

Where am I going wrong? I asked my tutor but he said, he doesn't really know, and only that Newtons second law like that would leave out the fact its rotational motion.

I thought Newton's law would suffice to describe the translational motion of the CM like that but apparently not, so why is this wrong?

• You are correct, the given answer is wrong. If there's no friction, there's no rolling... or more to the point, no angular speed change. Commented Jan 10, 2023 at 20:46
• Presumably, the text should speak of pure rolling without slipping, which is not equivalent to without friction. Commented Jan 11, 2023 at 6:35

The key word "pure rolling" means rolling without slipping. In other words, at all times the velocity of the material of the ball at the point of contact with the plane is 0. There is no slipping between the surface of the ball and the surface of the plane.

However, this "pure rolling" condition requires friction. You cannot have both "pure rolling" and "without friction". If the problem truly specifies both, then the problem is self-contradictory. However, you may have misread the problem. It may have specified "no rolling resistance" which is different from "without friction".

Newton's 2nd law applies because you must consider the net force acting on the ball, and that includes not only gravity but also the static friction that keeps the ball rolling. Friction is important here because the ball rotationally accelerates and thus there must be a net torque about the center of mass. This torque comes from the contact with the ramp.

Only if you disregard friction and the ball is sliding down a ramp, you will get the "expected" acceleration.

• There is something missing in your explanation. One should explain why the presence of torque is equivalent to a reduction of the center of mass acceleration. Commented Jan 11, 2023 at 6:33
• @GiorgioP - The torque isn't responsible for that, the force of friction is. Commented Jan 11, 2023 at 8:01

You can solve your task by starting with considering Newton's 2nd law in its rotational version: $$\sum \tau = I \alpha.$$
The key is to link the linear properties (which is what you are seeking, linear/translation acceleration) with the rotational properties, and for that you might as well need a geometric bond, such as $$a=r\alpha$$.