Initial Acceleration of a Spinning and Translating Disc

The friction direction varies across the contact surface, so the net force must be obtained by integrating the local kinetic-friction vectors.

\(a_{\mathrm{CM}} \simeq \mu g\,v_0/(\omega_0R)\), opposite to \(\mathbf v_0\)

Problem

A uniform disc of radius \(R\) rests with one flat face on a rough horizontal floor. The coefficient of friction is \(\mu\). The disc is given angular velocity \(\omega_0\) about its vertical central axis and, simultaneously, horizontal velocity \(v_0\), where \(v_0\ll R\omega_0\). Find a suitable expression for the initial acceleration of its centre of mass.

Velocity of a contact point

Coordinates and assumptions

Choose the translational velocity along \(+x\): \(\mathbf v_0=v_0\hat{\mathbf x}\).
Take \(\boldsymbol {\ omega}_0=\omega_0\hat{\mathbf z}\).
A point on the disc is labelled by polar coordinates \((r,\phi)\).
The normal pressure is uniform: \(p=Mg/(\pi R^2)\).
Every contact element is sliding initially, so kinetic friction opposes its local velocity.

Top view: local velocity is the vector sum of translation and rotation.

For the point \((r,\phi)\), the velocity relative to the floor is

\mathbf u =\mathbf v_0+\boldsymbol\omega_0\times\mathbf r =\bigl(v_0-\omega_0r\sin\phi\bigr)\hat{\mathbf x} +\omega_0r\cos\phi\,\hat{\mathbf y}.

Write the local friction force

The friction on an area element \(dA=r\,dr\,d\phi\) has magnitude \(\mu p\,dA\) and points opposite to \(\mathbf u\). Its \(x\)-component is therefore

dF_x=-\mu p\,r\,dr\,d\phi\, \frac{v_0-\omega_0r\sin\phi} {\sqrt{\omega_0^2r^2+v_0^2-2v_0\omega_0r\sin\phi}}.

The components perpendicular to \(\mathbf v_0\) cancel by symmetry, so the net friction—and hence the acceleration of the centre—is parallel or antiparallel to \(\mathbf v_0\).

Use \(v_0\ll \omega_0R\)

For a ring with \(r\gg v_0/\omega_0\), define \(\varepsilon=v_0/(\omega_0r)\). To first order in \(\varepsilon\),

\frac{u_x}{|\mathbf u|} =\frac{\varepsilon-\sin\phi} {\sqrt{1-2\varepsilon\sin\phi+\varepsilon^2}} \simeq-\sin\phi+\varepsilon\cos^2\phi.

Now integrate around a complete ring. The \(-\sin\phi\) contribution vanishes, whereas

\int_0^{2\pi}\cos^2\phi\,d\phi=\pi.

Thus the net horizontal friction is

\begin{aligned} F_x &\simeq-\mu p\int_0^R r\,dr \int_0^{2\pi}\frac{v_0}{\omega_0r}\cos^2\phi\,d\phi\\[4pt] &=-\mu p\,\frac{\pi Rv_0}{\omega_0}. \end{aligned}

About the centre

The expansion is not valid in the tiny region \(r\lesssim v_0/\omega_0\), but that region has area of order \((v_0/\omega_0)^2\). Its contribution is second order and does not change the leading result.

Acceleration of the centre of mass

With uniform pressure,

p=\frac{Mg}{\pi R^2}.

Substituting this in the force expression and dividing by \(M\),

Initial acceleration

\[ \boxed{ \mathbf a_{\mathrm{CM}} \simeq-\frac{\mu g}{\omega_0R}\,\mathbf v_0 } \]

Therefore, \(\displaystyle |a_{\mathrm{CM}}|\simeq\frac{\mu g v_0}{\omega_0R}\), directed opposite to the translational velocity.

Coefficient check. The angular integration contains \(\cos^2\phi\), whose average over a complete circle is \(1/2\). Replacing this angular factor by unity would double the result. The correct leading coefficient is therefore \(1\), not \(2\).

Alternate solution contributed by Sayan Chatterjee. 📄 Download Solution PDF