# Finding a Rotated Center of Mass

Suppose we have a stationary cylinder of radius $$R$$ on top of which is balanced a rectangular slab of thickness $$a$$. Suppose the slab is rotated (without slipping) about an angle $$\theta$$, so that the point of contact of the slab with the cylinder is rotated through an angle $$\theta$$. I want to find out the coordinates of the centroid of the rotated slab with respect to the coordinates of the centroid of the un-rotated slab. I have drawn the following figure:

I'm having trouble getting started, and would appreciate any help. I apologize for not including any of my own work on this problem, but it consists mostly in a lot of triangles that don't seem to lead anywhere. I am happy to include my work, in keeping with the spirit of this forum, if it could be relevant.

• @whatamidoing yes Commented Jun 24 at 12:08
• Note that the problem is essentially same if, instead of rotating, the slab is just moved the slab left by an amount $(R+a/2)\tan \theta$. In order to close the gap between these two, you just need to change the coordinate system. Commented Jun 24 at 12:12
• @MattiP. thanks for your help! I'm going to have to take a bit to figure out how to use it to answer my question. After the translation you mentioned, would the change in coordinates just amount to a rotation? And why is the translation by $(R+a/2)\tan{(\theta)}$ and not $R\tan{(\theta)}$? Commented Jun 24 at 12:24

We can find the displacement vector adding the following three vectors:

• from the center of the cylinder to the point of contact of the slab with the cylinder

$$\vec v_1 =(R\sin \theta, R\cos \theta)$$

• from the point of contact of the slab with the cylinder to the initial point of contact

$$\vec v_2 =(-R\theta\cos \theta, R\theta\sin \theta)$$

• from the initial point of contact to the center of mass

$$\vec v_3 =\left(\frac a2\sin \theta, \frac a2\cos \theta\right)$$

and then subtracting the initial position vector of the center of mass with respect to the center of the cylinder

$$\vec v_4 =\left(0, R+\frac a2\right)$$

that is

$$\vec v_1+\vec v_2+\vec v_3-\vec v_4 =\left(\left(R+\frac a2\right)\sin \theta-R\theta\cos \theta, \left(R+\frac a2\right)(\cos \theta-1)+R\theta\sin \theta\right)$$

• Thank you for your thorough response! I just have one question. In deriving the expression for $\mathbf{v}_{2}$, you have that its length is $R\theta$. Is this exact? I thought $R\theta$ was the length of arc of the circle subtended by the angle $\theta$, but is it also the length of $\mathbf{v}_{2}$? Commented Jun 25 at 13:18
• @GeorgyZhukov Yes of course, because the slab rotates without slipping, the length is the same.
– user
Commented Jun 25 at 13:24
• Ah, of course. Thank you very much. Commented Jun 25 at 13:53

This is a two-dimensional problem. Take the origin of the $$xy$$ plane to be the center of the circle (the cross-section of the cylidner). Then the initial COM (Center of Mass) location is at

$$P_1 = \begin{bmatrix} 0 \\ r + \dfrac{a}{2} \end{bmatrix}$$

Now after rolling the slab, the point of contact $$Q$$ is easy to calculate

$$Q = \begin{bmatrix} r \sin \theta \\ r \cos \theta \end{bmatrix}$$

The distance $$d$$ covered on both the cylinder and the slab is

$$d = r \theta$$

And the rotation matrix describing the orientation of the slab local frame with respect to the cylinder frame is

$$R = \begin{bmatrix} \cos \theta && \sin \theta \\ - \sin \theta && \cos \theta \end{bmatrix}$$

The frame of the slab relation to the cylinder frame (which is centered at the center of the circular cross section) is

$$p = C + R p'$$

where $$p$$ is the world coordinate position vector, and $$p'$$ is the position vector relative the frame attached to the slab, and $$C$$ is the origin of this frame, which is the centroid.

At the point of contact, $$p' = ( d , - \dfrac{a}{2} )$$, and $$p = (r \sin \theta, r \cos \theta)$$

Therefore,

$$C = p - R p' = \begin{bmatrix} r \sin \theta \\ r \cos \theta \end{bmatrix} - \begin{bmatrix} \cos \theta && \sin \theta \\ - \sin \theta && \cos \theta \end{bmatrix} \begin{bmatrix} r \theta \\ - \dfrac{a}{2} \end{bmatrix}$$

And this evaluates to

$$C = \begin{bmatrix} r \sin \theta - r \theta \cos \theta + \dfrac{1}{2} a \sin \theta \\ r \cos \theta + r \theta \sin \theta + \dfrac{1}{2} a \cos \theta \end{bmatrix}$$

Subtracting the coordinates of the un-rotated slab gives the relative coordinates (relative displacement)

$$\Delta C = \begin{bmatrix} r \sin \theta - r \theta \cos \theta + \dfrac{1}{2} a \sin \theta \\ r \theta \sin \theta + (\dfrac{1}{2} a + r) (\cos \theta -1 ) \end{bmatrix}$$

HINT:

Two things to be considered here while finding locus of C:

1. The slab edge $$P'T'$$ is like a taut string and when it rolls, the initial point displaces $$P\to P', T\to T', S\to S'$$ while describing an involute (red). "String" length equals arc length $$P'T'= R \theta$$.

The base of the triangle $$C'P'T'=P'T'$$is variable, the length of "taut string".

1. Consider the length of the "unwound string" line segment $$P'T'$$ as hypotenuse to find Center of Mass by $$(x,y)$$ projections (green) directly.

In this rough sketch it should be noted that the coincident point $$(P,T,Q,S)$$ disassociated into four separate points $$(P',T',Q',S')$$

$$(x'',y'')=$$

$$R(\sin \theta - \theta \cos \theta ) + a \sin \theta/2 +R (\cos \theta + \theta \sin \theta )+ a \cos \theta/2$$

We may call it as a modified circle involute.

Another very rough sketch of C locus when slab rolls on the circle $$(R=1, a=0.4)$$ :