Water in a rotating container of radius \(50 \mathrm{~mm}\) is \(30 \mathrm{~mm}\) lower in the centre than at the edge. Find the angular velocity of the container.

The forces acting on the water in the container are the gravitational force and the centripetal force. The gravitational force acts vertically downwards, while the centripetal force acts horizontally towards the center of the container as it rotates. The centripetal force is given by the equation \(F_{c} = m \omega^2 r\) where \(F_{c}\) is the centripetal force, \(m\) is the mass of the water, \(\omega\) is the angular velocity, and \(r\) is the distance from the center of the container.

Due to the rotational motion of the container, the water's shape is determined by an equilibrium between centrifugal force and gravitational force. Let \(h\) be the difference between the height at the edge and the height in the center, then we can relate gravitational force with centrifugal force as follows: \(\rho g h = \rho (\omega r)^2/2\) Where \(\rho\) is the water density, \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{ms^{-2}}\)), \(h\) is the height difference, which is \(30 \mathrm{mm}\), and \(r\) is the radius of the container, which is \(50 \mathrm{mm}\). Note that we've divided by 2 on the right side of the equation because we're calculating the difference in heights at the center and the edge.

Now, we can solve the equation for angular velocity: \(\omega^2 = \frac{2gh}{r^2}\) Substituting the known values, we have: \(\omega^2 = \frac{2(9.8 \mathrm{ms^{-2}})(0.03 \mathrm{m})}{(0.05 \mathrm{m})^2}\) Solving for \(\omega\): \(\omega = \sqrt{\frac{2(9.8)(0.03)}{(0.05)^2}} = \sqrt{23.52} \approx 4.85 \mathrm{rad/s}\) Therefore, the angular velocity of the container is approximately \(4.85 \mathrm{rad/s}\).