# Problem Restatement A drum with a central hole in its lid is filled with liquid up to a height \( d \), where the has density \( \rho \). The drum is spun at angular velocity \( \omega \) (you wrote \(\alpha\), but the answer uses \(\omega\) as standard for angular velocity). What is the **inner radius \( r_i \) of the liquid at the lid** (i.e., the radius of the liquid's surface at the lid), once the liquid has redistributed due to rotation? The answer is given: \[ r_i = \left[ \frac{4g r_o^2 (d_o - d)}{\omega^2} \right]^{1/4} \] where \( r_o \) is the drum's outer radius, \( d_o \) is the initial liquid depth (before spinning), and \( d \) is the depth at the center after spinning. # Step-by-Step Solution ## 1. **Understanding the Physical Situation** - When the drum spins, the free surface of the liquid forms a paraboloid due to the combination of gravity and centrifugal effects. - At the center, the liquid surface dips down to a new height \( d \) (less than the initial height \( d_o \)). - At the lid (at \( r = r_i \)), the liquid just touches the lid (i.e., the surface reaches up to the lid's hole). ## 2. **Equation of the Rotating Liquid Surface** The surface profile of a rotating liquid is: \[ z(r) = z() + \frac{\omega^2 r^2}{2g} \] where: - \( z(r) \): height of liquid surface at radius \( r \) - \( z() \): height at the center (here, \( d \)) - \( g \): acceleration due to gravity - \( \omega \): angular velocity So, \[ z(r) = d + \frac{\omega^2 r^2}{2g} \] ## 3. **At the Inner Radius at the Lid** At the lid's hole (\( r = r_i \)), the liquid just reaches up to the lid (height zero above the hole): \[ z(r_i) = = d + \frac{\omega^2 r_i^2}{2g} \] So, \[ = d + \frac{\omega^2 r_i^2}{2g} \implies r_i^2 = -\frac{2g d}{\omega^2} \] But since \( d \) is taken positive **below the lid**, we need to set our reference: let the lid be \( z = \), then the liquid surface at center is at \( z = -d \): \[ z(r) = -d + \frac{\omega^2 r^2}{2g} \] At the inner edge (\( r = r_i \)), \( z(r_i) = \): \[ = -d + \frac{\omega^2 r_i^2}{2g} \implies \frac{\omega^2 r_i^2}{2g} = d \implies r_i^2 = \frac{2g d}{\omega^2} \] ## 4. **Conservation of Liquid Volume** The total volume remains constant before and after spinning. - **Initial volume:** (cylinder of radius \( r_o \), height \( d_o \)) \[ V_{\text{initial}} = \pi r_o^2 d_o \] - **After spinning:** Volume is the volume under the paraboloid (with inner radius \( r_i \), outer radius \( r_o \)), up to the lid (height zero at \( r_i \)), and depth \( -d \) at the center. The volume under the paraboloid from \( r = r_i \) to \( r = r_o \) is: \[ V_{\text{final}} = \int_{r_i}^{r_o} 2\pi r\, z(r)\, dr \] Recall, \( z(r) = -d + \frac{\omega^2 r^2}{2g} \): \[ V_{\text{final}} = \int_{r_i}^{r_o} 2\pi r \left[ -d + \frac{\omega^2 r^2}{2g} \right] dr \] \[ = 2\pi \int_{r_i}^{r_o} \left( -d r + \frac{\omega^2 r^3}{2g} \right) dr \] \[ = 2\pi \left[ -\frac{d}{2}(r_o^2 - r_i^2) + \frac{\omega^2}{8g}(r_o^4 - r_i^4) \right] \] ## 5. **Equate Volumes and Solve for \( r_i \)** Set \( V_{\text{initial}} = V_{\text{final}} \): \[ \pi r_o^2 d_o = 2\pi \left[ -\frac{d}{2}(r_o^2 - r_i^2) + \frac{\omega^2}{8g}(r_o^4 - r_i^4) \right] \] Divide by \( \pi \): \[ r_o^2 d_o = 2 \left[ -\frac{d}{2}(r_o^2 - r_i^2) + \frac{\omega^2}{8g}(r_o^4 - r_i^4) \right] \] Multiply both sides by 1/2: \[ \frac{r_o^2 d_o}{2} = -\frac{d}{2}(r_o^2 - r_i^2) + \frac{\omega^2}{8g}(r_o^4 - r_i^4) \] Bring all terms to one side: \[ \frac{r_o^2 d_o}{2} + \frac{d}{2}(r_o^2 - r_i^2) - \frac{\omega^2}{8g}(r_o^4 - r_i^4) = \] Simplify \( \frac{d}{2}(r_o^2 - r_i^2) \): \[ \frac{r_o^2 d_o}{2} + \frac{d r_o^2}{2} - \frac{d r_i^2}{2} - \frac{\omega^2}{8g}(r_o^4 - r_i^4) = \] \[ \Rightarrow \frac{r_o^2}{2}(d_o + d) - \frac{d r_i^2}{2} - \frac{\omega^2}{8g}(r_o^4 - r_i^4) = \] Let’s bring terms with \( r_i \) to one side: \[ \frac{r_o^2}{2}(d_o + d) - \frac{\omega^2}{8g} r_o^4 = \frac{d r_i^2}{2} - \frac{\omega^2}{8g} r_i^4 \] \[ \frac{r_o^2}{2}(d_o + d) - \frac{\omega^2}{8g} r_o^4 = \frac{1}{2} d r_i^2 - \frac{\omega^2}{8g} r_i^4 \] \[ \left[ \frac{1}{2} d r_i^2 - \frac{\omega^2}{8g} r_i^4 \right] = \frac{r_o^2}{2}(d_o + d) - \frac{\omega^2}{8g} r_o^4 \] Let’s factor \( r_i^2 \): \[ \frac{1}{2} d r_i^2 - \frac{\omega^2}{8g} r_i^4 = \frac{r_o^2}{2}(d_o + d) - \frac{\omega^2}{8g} r_o^4 \] Bring all terms to one side: \[ \frac{1}{2} d r_i^2 - \frac{\omega^2}{8g} r_i^4 - \frac{r_o^2}{2}(d_o + d) + \frac{\omega^2}{8g} r_o^4 = \] \[ \frac{1}{2} d r_i^2 - \frac{\omega^2}{8g} r_i^4 = \frac{r_o^2}{2}(d_o + d) - \frac{\omega^2}{8g} r_o^4 \] Alternatively, moving terms to one side: \[ \frac{1}{2} d r_i^2 - \frac{\omega^2}{8g} r_i^4 = \frac{r_o^2}{2}(d_o + d) - \frac{\omega^2}{8g} r_o^4 \] But we can also recognize that the initial volume can be written as: \[ V = \int_{r_i}^{r_o} 2\pi r z(r) dr = \pi r_o^2 d_o \] But for the purposes of the given answer, let's try a shortcut. Note from the surface equation earlier (step 3): \[ r_i^2 = \frac{2g d}{\omega^2} \] So, \[ d = \frac{\omega^2 r_i^2}{2g} \] Let's substitute this for \( d \) in the volume equation to eliminate \( d \): So, from the earlier volume conservation, \[ \pi r_o^2 d_o = 2\pi \left[ -\frac{d}{2}(r_o^2 - r_i^2) + \frac{\omega^2}{8g}(r_o^4 - r_i^4) \right] \] \[ r_o^2 d_o = -d(r_o^2 - r_i^2) + \frac{\omega^2}{4g}(r_o^4 - r_i^4) \] Now, substitute \( d = \frac{\omega^2 r_i^2}{2g} \): \[ r_o^2 d_o = -\left( \frac{\omega^2 r_i^2}{2g} \right)(r_o^2 - r_i^2) + \frac{\omega^2}{4g} (r_o^4 - r_i^4) \] \[ r_o^2 d_o = -\frac{\omega^2 r_i^2 (r_o^2 - r_i^2)}{2g} + \frac{\omega^2}{4g} (r_o^4 - r_i^4) \] Let’s write \( r_o^4 - r_i^4 = (r_o^2 - r_i^2)(r_o^2 + r_i^2) \): \[ r_o^2 d_o = -\frac{\omega^2 r_i^2 (r_o^2 - r_i^2)}{2g} + \frac{\omega^2}{4g} (r_o^2 - r_i^2)(r_o^2 + r_i^2) \] So factor \( r_o^2 - r_i^2 \): \[ r_o^2 d_o = (r_o^2 - r_i^2) \left[ -\frac{\omega^2 r_i^2}{2g} + \frac{\omega^2}{4g}(r_o^2 + r_i^2) \right] \] \[ = (r_o^2 - r_i^2) \left[ \frac{\omega^2}{4g}(r_o^2 + r_i^2) - \frac{\omega^2 r_i^2}{2g} \right] \] \[ = (r_o^2 - r_i^2) \left[ \frac{\omega^2}{4g}(r_o^2 + r_i^2 - 2 r_i^2) \right] \] \[ = (r_o^2 - r_i^2) \left[ \frac{\omega^2}{4g}(r_o^2 - r_i^2) \right] \] Thus, \[ r_o^2 d_o = \frac{\omega^2}{4g}(r_o^2 - r_i^2)^2 \] Now solve for \( r_i^2 \): \[ (r_o^2 - r_i^2)^2 = \frac{4g r_o^2 d_o}{\omega^2} \] \[ r_o^2 - r_i^2 = \left( \frac{4g r_o^2 d_o}{\omega^2} \right)^{1/2} \] \[ r_i^2 = r_o^2 - \left( \frac{4g r_o^2 d_o}{\omega^2} \right)^{1/2} \] But the given answer is in terms of \( d_o - d \): Recall that \( r_i^2 = \frac{2g d}{\omega^2} \implies d = \frac{\omega^2 r_i^2}{2g} \), so \( d_o - d = d_o - \frac{\omega^2 r_i^2}{2g} \). From above: \[ (r_o^2 - r_i^2)^2 = \frac{4g r_o^2 d_o}{\omega^2} \] But if the liquid at the center is at depth \( d \) and at the rim the