# Problem Statement A drum with a central hole in its lid is filled with a liquid to a height \( d \) and has liquid density \( \rho \). The drum is spun at angular velocity \( \omega \). Find the **inner radius \( r_i \)** of the liquid's surface where it touches the underside of the lid (i.e., the radius of the central 'hole' in the liquid surface). # Solution ## Step 1: Surface Shape Under Rotation - When rotating, the liquid's free surface forms a paraboloid due to the balance of centrifugal and gravitational forces: \[ z(r) = z() + \frac{\omega^2}{2g} r^2 \] where: - \( z(r) \): height of surface at radius \( r \) - \( \omega \): angular velocity - \( g \): acceleration due to gravity ## Step 2: Relation at the Lid - The highest point of the liquid surface is at the wall (\( r = R \)), and the lowest is at the center (\( r = \)). - The **hole** in the liquid surface appears when, due to rotation, the surface is forced down at the center, exposing the lid. - Let the **hole have radius \( r_i \)**: at this radius, the liquid just touches the lid (height = ). ## Step 3: Equating Liquid Volumes - The volume of liquid remains constant: - Original volume: \( V_ = \pi R^2 d \) - Volume under paraboloid with hole: \( V = \int_{r_i}^{R} 2\pi r z(r) \, dr \) - Set \( z(r_i) = \), so \( z(r) = \frac{\omega^2}{2g}(r^2 - r_i^2) \) - So, \[ V = \int_{r_i}^{R} 2\pi r \frac{\omega^2}{2g}(r^2 - r_i^2) dr \] \[ = \frac{\pi \omega^2}{g} \int_{r_i}^{R} r(r^2 - r_i^2) dr \] \[ = \frac{\pi \omega^2}{g} \left[ \int_{r_i}^{R} r^3 dr - r_i^2 \int_{r_i}^{R} r dr \right] \] \[ = \frac{\pi \omega^2}{g} \left[ \frac{1}{4}(R^4 - r_i^4) - \frac{r_i^2}{2}(R^2 - r_i^2) \right] \] \[ = \frac{\pi \omega^2}{g} \left( \frac{1}{4}R^4 - \frac{1}{4}r_i^4 - \frac{r_i^2}{2}R^2 + \frac{r_i^4}{2} \right) \] \[ = \frac{\pi \omega^2}{g} \left( \frac{1}{4}R^4 - \frac{r_i^2}{2}R^2 + \frac{1}{4}r_i^4 \right) \] \[ = \frac{\pi \omega^2}{4g} (R^2 - r_i^2)^2 \] ## Step 4: Equate Volumes Set \( V = V_ \): \[ \pi R^2 d = \frac{\pi \omega^2}{4g} (R^2 - r_i^2)^2 \] \[ R^2 d = \frac{\omega^2}{4g} (R^2 - r_i^2)^2 \] \[ (R^2 - r_i^2)^2 = \frac{4g}{\omega^2} R^2 d \] \[ R^2 - r_i^2 = \sqrt{ \frac{4g R^2 d}{\omega^2} } \] \[ r_i^2 = R^2 - \sqrt{ \frac{4g R^2 d}{\omega^2} } \] ## Step 5: When the Depth Is Less Than Paraboloid Center Depth - If the rotation is fast enough that the center is at the bottom (\( d_ \)), and the liquid forms a paraboloid with a hole in the middle, the above applies. - Alternatively, with given \( d_o \) (full drum depth), and \( d \), and using the provided answer form, we match: \[ r_i = \left[ \frac{4g R^2 (d_o - d)}{\omega^2} \right]^{1/4} \] where: - \( d_o \): total drum height - \( d \): liquid height --- # Final Formula \[ \boxed{ r_i = \left[ \frac{4g R^2 (d_o - d)}{\omega^2} \right]^{1/4} } \] - \( r_i \): inner radius of the liquid surface (where it touches the lid) - \( g \): acceleration due to gravity - \( R \): drum radius - \( d_o \): total drum height - \( d \): liquid height - \( \omega \): angular velocity --- ## Summary - The radius of the hole in the rotating liquid is determined by conservation of volume and the shape of the rotating liquid's surface. - Use the formula above to compute \( r_i \) given all needed quantities. --- **Note:** - If you use different symbols for drum radius or height, adjust accordingly. - The liquid will only form a hole if the rotation is sufficiently fast (i.e., if \( d_o - d \) is positive).