# Cylinder rolling on inclined plane

Let’s consider a rolling cylinder on an inclined plane with some friction. We knew that $$\alpha R=a$$, where $$\alpha$$ is the angular acceleration of each constituent particle and $$R$$ is the radius of disc, so $$a$$ will be the acceleration of each constituent particle during the rotation and depends on their distance from the center of cylinder.

My question here is how can the acceleration $$a=\alpha R$$ be equal to the acceleration of center of mass down the inclined plane?

In rotational dynamics, $$a=\alpha r$$ represents the relationship between angular acceleration and tangential acceleration of a point particle.

In rolling motion, we can analyze it in two motions; translational and rotational. Let $$v$$ be the translational velocity of com and $$\omega$$ be the angular velocity. If we consider the point that touches the plane, it has $$v$$ velocity to the forward and $$v'=\omega r$$ tangential velocity to the backward. So there net velocity is $$v-v'$$. If this is a rolling without slipping motion, that particle has zero net velocity. Therefore \begin{align}v-v' &=0\\v-\omega r &=0\\v &=\omega r\end{align} If you differentiate this w.r.t. time you will get $$a=\alpha r$$ whereas $$a=\frac{dv}{dt}$$ and $$\alpha=\frac{d\omega}{dt}$$.

• so if I consider the top particle it will have a net velocity of v+v(prime), am I right? Aug 18, 2021 at 6:54
• Yes, which mean v+v=2v velocity as v=v' from that condition.
– ACB
Aug 18, 2021 at 6:55
• @Eugene , is this answer acceptable?
– ACB
Aug 18, 2021 at 7:34
• Yes, but I still have a question here, if slipping occurs will I have the velocity of center of mass greater or equal than angular velocity times the radius? Aug 19, 2021 at 2:25
• Thank you very much! Aug 19, 2021 at 8:23

You are right -- each point on the cylinder experiences a different acceleration.

When dealing with rotational motion, it's common to break things up into rotation about the center of mass ($$\alpha$$), and the translation of the center of mass ($$a$$).

The $$a$$ you refer to is the acceleration of the center of mass (one point, the center of your cylinder if it's uniform).

Next, $$\alpha$$ represents the angular acceleration which describes the motion of the edges of the cylinder and how their rotation around the cylinder's center of mass is being accelerated.

Lastly, note that $$a=\alpha R$$ is a scalar equation, which simply relates the magnitudes of the accelerations.

• But why the product between angular acceleration and radius gives the acceleration of center of mass? Aug 18, 2021 at 6:10
• that is the condition for rolling without slipping @Eugene Aug 18, 2021 at 6:13
• Sorry , I have a bit trouble understanding why it is the condition for rolling without slipping…… Aug 18, 2021 at 6:33