International Physics Online Olympiad 2014

Consider a uniform semi-circular disk as shown in the diagram. To compute the X and Y coordinates of the COM of this uniform semi-circular disk, the following two relations are used. Let the mass per unit area be $\sigma$ .

$X_C = \frac{\int_{A} x \ dm}{\int_{A} \ dm} = \frac{\int_{0}^{\pi}\int_{0}^{R} (R + r\cos{\theta})(\sigma \ r \ dr \ d\theta)}{\int_{0}^{\pi}\int_{0}^{R}\sigma \ r \ dr \ d\theta} =R$ $Y_C = \frac{\int_{A} y \ dm}{\int_{A} \ dm} = \frac{\int_{0}^{\pi}\int_{0}^{R} (r\sin{\theta})(\sigma \ r \ dr \ d\theta)}{\int_{0}^{\pi}\int_{0}^{R}\sigma \ r \ dr \ d\theta} =\frac{4R}{3\pi}$

Having found the COM coordinates, imagine that the disk is rotated about the Z-axis such that the point on it coincident with the origin is fixed. If it is rotated anti-clockwise by a small angle, it is expected to return to its original position due to the component of weight that brings it back down.

One can see that the net unbalanced torque about the origin in the clockwise direction brings it back to its initial orientation. The only way that the disk can tip over is when the angle is increased so that the weight is to the left of the reaction force. The point at which the disk just tips over is when the weight and reaction forces are aligned. This happens when the COM intersects the Y-axis. In other words, the disk must be rotated by an angle so that the COM lies exactly on the Y axis. Let the initial position vector of the COM be:

$\vec{r} = X_C \ \hat{i} + Y_C \ \hat{j}$

The angle this vector makes with the horizontal is:

$\alpha = \arctan\left(\frac{Y_C}{X_C}\right)= \arctan\left(\frac{4}{3\pi}\right)$

Therefore the angle by which the disk must be rotated is:

$\gamma = \frac{\pi}{2} - \alpha = \frac{\pi}{2} -\arctan\left(\frac{4}{3\pi}\right)$

$\boxed{\gamma \approx 67^{\circ}}$