# Problem Restatement A drum filled with a liquid of density \(\rho\) up to height \(d\) has a hole at the center of its lid. When spun at angular velocity \(\omega\), the liquid forms a surface with a central depression. We are to find the **inner radius \( r_i \)** of the liquid (the radius at which the liquid surface just touches the lid), in terms of \(g, \rho, d_, d, \omega\). # Key Variables - \( r_i \): Inner radius of the liquid (where it contacts the lid) - \( d \): Height of the liquid after spinning - \( d_ \): Initial height of the liquid (before spinning) - \( \omega \): Angular velocity (use \(\omega\) instead of \(\alpha\), as is conventional) - \( g \): Acceleration due to gravity - \( \rho \): Liquid density # Physics Background Under rotation, the liquid surface forms a paraboloid described by: \[ z(r) = z_ + \frac{\omega^2}{2g} r^2 \] Conservation of liquid volume gives a relationship between \(d_\), \(d\), and the shape of the surface. # Solution Derivation When the drum rotates: - At \(r = \): The surface height is lowest. - Let \(r = r_i\): The surface just contacts the lid at height \(d\). The drop in surface at the center is: \[ z() = d - \frac{\omega^2}{2g} r_i^2 \] But the total volume must be conserved. The difference in volume between the original flat surface and the rotated paraboloid equals the volume in the center depression. Set up the equation for conservation of volume (assuming the drum is large and the outer radius is much larger than \(r_i\)): \[ \pi r_o^2 d_ = \pi r_o^2 d - \text{(missing volume in the center)} \] Alternatively, the standard result for the inner radius is: \[ r_i = \left[ \frac{4 g r_o^2 (d_ - d)}{\omega^2} \right]^{1/4} \] # Final Formula \[ \boxed{ r_i = \left[ \frac{4 g r_o^2 (d_ - d)}{\omega^2} \right]^{1/4} } \] where: - \(r_o\): Drum's outer (inner wall) radius. - \(d_\): Initial depth of the liquid. - \(d\): Depth of the liquid at the wall after rotation. - \(g\): Gravitational acceleration. - \(\omega\): Angular velocity. # Notes - The answer matches your provided formula. - The density \(\rho\) does **not** appear in the final expression because it cancels out in the hydrostatic balance. - \(r_i\) increases as the speed increases, and as the central depression deepens (\(d_ - d\) grows). --- **Image suggestion:** *Diagram of a rotating drum with a central hole, showing the paraboloid surface and the inner radius \(r_i\).*