IPhO 2008 · Theory problem 1

Water-Powered Rice-Pounding Mortar

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A. Introduction

Rice is the main staple food of most people in Vietnam. To make white rice from paddy rice, one needs separate of the husk (a process called "hulling") and separate the bran layer ("milling"). The hilly parts of northern Vietnam are abundant with water streams, and people living there use water-powered rice-pounding mortar for bran layer separation. Figure 1 shows one of such mortars, Figure 2 shows how it works.

B. Design and operation

1. Design.

The rice-pounding mortar shown in Figure 1 has the following parts:

The mortar, basically a wooden container for rice.

The lever, which is a tree trunk with one larger end and one smaller end. It can rotate around a horizontal axis. A pestle is attached perpendicularly to the lever at the smaller end. The length of the pestle is such that it touches the rice in the mortar when the lever lies horizontally. The larger end of the lever is carved hollow to form a bucket. The shape of the bucket is crucial for the mortar's operation.

2. Modes of operation

The mortar has two modes.

Working mode. In this mode, the mortar goes through an operation cycle illustrated in Figure 2.

The rice-pounding function comes from the work that is transferred from the pestle to the rice during stage f) of Figure 2. If, for some reason, the pestle never touches the rice, we say that the mortar is not working.

Rest mode with the lever lifted up. During stage c) of the operation cycle (Figure 2), as the tilt angle \(\alpha\) increases, the amount of water in the bucket decreases. At one particular moment in time, the amount of water is just enough to counterbalance the weight of the lever. Denote the tilting angle at this instant by \(\beta\). If the lever is kept at angle \(\beta\) and the initial angular velocity is zero, then the lever will remain at this position forever. This is the rest mode with the lever lifted up. The stability of this position depends on the flow rate of water into the bucket, \(\Phi\). If \(\Phi\) exceeds some value \(\Phi_2\), then this rest mode is stable, and the mortar cannot be in the working mode.

In other words, \(\Phi_2\) is the minimal flow rate for the mortar not to work.

Figure 1

Figure 1
A water-powered rice-pounding mortar

OPERATION CYCLE OF A WATER-POWERED RICE-POUNDING MORTAR

Figure 2

Figure 2

a) At the beginning there is no water in the bucket, the pestle rests on the mortar. Water flows into the bucket with a small rate, but for some time the lever remains in the horizontal position.

b) At some moment the amount of water is enough to lift the lever up. Due to the tilt, water rushes to the farther side of the bucket, tilting the lever more quickly.

Water starts to flow out at \(\alpha=\alpha_1\).

c) As the angle \(\alpha\) increases, water starts to flow out. At some particular tilt angle, \(\alpha=\beta\), the total torque is zero.

d) \(\alpha\) continues increasing, water continues to flow out until no water remains in the bucket.

e) \(\alpha\) keeps increasing because of inertia. Due to the shape of the bucket, water falls into the bucket but immediately flows out. The inertial motion of the lever continues until \(\alpha\) reaches the maximal value \(\alpha_0\).

f) With no water in the bucket, the weight of the lever pulls it back to the initial horizontal position. The pestle gives the mortar (with rice inside) a pound and a new cycle begins.

C. The problem

Consider a water-powered rice-pounding mortar with the following parameters (Figure 3)

The mass of the lever (including the pestle but without water) is \(M=30\ \mathrm{kg}\),

The center of mass of the lever is G. The lever rotates around the axis T (projected onto the point T on the figure).

The moment of inertia of the lever around T is \(I=12\ \mathrm{kg\cdot m^2}\).

When there is water in the bucket, the mass of water is denoted as \(m\), the center of mass of the water body is denoted as N.

The tilt angle of the lever with respect to the horizontal axis is \(\alpha\).

The main length measurements of the mortar and the bucket are as in Figure 3.

Neglect friction at the rotation axis and the force due to water falling onto the bucket. In this problem, we make an approximation that the water surface is always horizontal.

Figure 3

Figure 3 Design and dimensions of the rice-pounding mortar

1. The structure of the mortar

At the beginning, the bucket is empty, and the lever lies horizontally. Then water flows into the bucket until the lever starts rotating. The amount of water in the bucket at this moment is \(m=1.0\ \mathrm{kg}\).

1.1. Determine the distance from the center of mass G of the lever to the rotation axis T. It is known that GT is horizontal when the bucket is empty.

1.2. Water starts flowing out of the bucket when the angle between the lever and the horizontal axis reaches \(\alpha_1\). The bucket is completely empty when this angle is \(\alpha_2\).

Determine \(\alpha_1\) and \(\alpha_2\).

1.3. Let \(\mu(\alpha)\) be the total torque (relative to the axis T) which comes from the weight of the lever and the water in the bucket. \(\mu(\alpha)\) is zero when \(\alpha=\beta\). Determine \(\beta\) and the mass \(m_1\) of water in the bucket at this instant.

2. Parameters of the working mode

Let water flow into the bucket with a flow rate \(\Phi\) which is constant and small. The amount of water flowing into the bucket when the lever is in motion is negligible. In this part, neglect the change of the moment of inertia during the working cycle.

2.1. Sketch a graph of the torque \(\mu\) as a function of the angle \(\alpha\), \(\mu(\alpha)\), during one operation cycle. Write down explicitly the values of \(\mu(\alpha)\) at angle \(\alpha_1\), \(\alpha_2\), and \(\alpha=0\).

2.2. From the graph found in section 2.1., discuss and give the geometric interpretation of the value of the total energy \(W_{\mathrm{total}}\) produced by \(\mu(\alpha)\) and the work \(W_{\mathrm{pounding}}\) that is transferred from the pestle to the rice.

2.3. From the graph representing \(\mu\) versus \(\alpha\), estimate \(\alpha_0\) and \(W_{\mathrm{pounding}}\) (assume the kinetic energy of water flowing into the bucket and out of the bucket is negligible.) You may replace curve lines by zigzag lines, if it simplifies the calculation.

3. The rest mode

Let water flow into the bucket with a constant rate \(\Phi\), but one cannot neglect the amount of water flowing into the bucket during the motion of the lever.

3.1. Assuming the bucket is always overflown with water,

3.1.1. Sketch a graph of the torque \(\mu\) as a function of the angle \(\alpha\) in the vicinity of \(\alpha=\beta\). To which kind of equilibrium does the position \(\alpha=\beta\) of the lever belong?

3.1.2. Find the analytic form of the torque \(\mu(\alpha)\) as a function of \(\Delta\alpha\) when \(\alpha=\beta+\Delta\alpha\), and \(\Delta\alpha\) is small.

3.1.3. Write down the equation of motion of the lever, which moves with zero initial velocity from the position \(\alpha=\beta+\Delta\alpha\) (\(\Delta\alpha\) is small). Show that the motion is, with good accuracy, harmonic oscillation. Compute the period \(\tau\).

3.2. At a given \(\Phi\), the bucket is overflown with water at all times only if the lever moves sufficiently slowly. There is an upper limit on the amplitude of harmonic oscillation, which depends on \(\Phi\). Determine the minimal value \(\Phi_1\) of \(\Phi\) (in kg/s) so that the lever can make a harmonic oscillator motion with amplitude \(1^\circ\).

3.3. Assume that \(\Phi\) is sufficiently large so that during the free motion of the lever when the tilting angle decreases from \(\alpha_2\) to \(\alpha_1\) the bucket is always overflown with water. However, if \(\Phi\) is too large the mortar cannot operate. Assuming that the motion of the lever is that of a harmonic oscillator, estimate the minimal flow rate \(\Phi_2\) for the rice-pounding mortar to not work.

Official English text of the problem.

Answers

Positive torque denotes increasing \(\alpha\). Numerical results use \(\rho=1000\ \mathrm{kg\,m^{-3}}\) and \(g=9.8\ \mathrm{m\,s^{-2}}\).

  • 1.1. \(\boxed{GT=1.57\ \mathrm{cm}}\).
  • 1.2. \(\boxed{\alpha_1=20.62^\circ,\quad \alpha_2=30^\circ}\).
  • 1.3. \(\boxed{\beta=23.57^\circ,\quad m_1=0.605\ \mathrm{kg}}\).
  • 2.1. On the rising stroke, \[ \boxed{\mu(0)=0,\quad \mu(\alpha_1)=+2.69\ \mathrm{N\,m}, \quad \mu(\alpha_2)=-4.00\ \mathrm{N\,m}}. \] The empty return branch is \(\boxed{\mu=-4.62\cos\alpha\ \mathrm{N\,m}}\), ending at \(-4.62\ \mathrm{N\,m}\). Filling at \(\alpha=0\) connects this value to zero.
  • 2.2. \(W_{\mathrm{total}}=\oint\mu\,d\alpha\) is the area enclosed by the cycle. The pounding work is \[ \boxed{W_{\mathrm{pounding}}=Mg(GT)\sin\alpha_0}, \] equal to \(W_{\mathrm{total}}\) when the kinetic energy carried away by water is neglected.
  • 2.3. \(\boxed{\alpha_0\approx37.5^\circ,\quad W_{\mathrm{pounding}}\approx2.8\ \mathrm{J}}\).
  • 3.1.1. Stable equilibrium; the torque crosses zero with negative slope at \(\beta\).
  • 3.1.2. With \(\Delta\alpha\) in radians, \[ \boxed{\mu(\beta+\Delta\alpha)\approx -(45.9\ \mathrm{N\,m\,rad^{-1}})\Delta\alpha}. \]
  • 3.1.3. \(\boxed{\alpha(t)=\beta+\Delta\alpha\cos(2\pi t/\tau)}\), with \(\boxed{\tau\approx3.26\ \mathrm{s}}\).
  • 3.2. \(\boxed{\Phi_1\approx0.23\ \mathrm{kg\,s^{-1}}}\).
  • 3.3. Using the limiting amplitude \(\beta-\alpha_1\approx2.95^\circ\), \[ \boxed{\Phi_2\approx0.7\ \mathrm{kg\,s^{-1}}}. \] The equilibrium is \(\beta\); \(\alpha_1\) is the lower turning point used in this estimate.

Solution

Write \(m_0=1.0\ \mathrm{kg}\) for the water charge that initiates motion, and \(r=GT\). From Figure 3, \[ a=0.20\ \mathrm m,\quad L=0.74\ \mathrm m,\quad h=0.12\ \mathrm m,\quad b=0.15\ \mathrm m,\quad \gamma=30^\circ. \] The pivot is \(d=0.08\ \mathrm m\) below the bucket’s upper edge. We use the standard values \(\rho=1000\ \mathrm{kg\,m^{-3}}\) and \(g=9.8\ \mathrm{m\,s^{-2}}\), which the problem does not specify.

Useful abbreviations are \[ q=\cot\gamma=\sqrt3,\qquad R=a+L=0.94\ \mathrm m,\qquad C=\frac{\rho b h^2}{2}=1.08\ \mathrm{kg}. \]

1.1. Position of the lever’s centre of mass

The bucket’s flat bottom has length \[ \ell=L-h\cot\gamma=0.532154\ \mathrm m. \]

Let \(\eta\) be the initial water depth above this bottom. The longitudinal water cross-section is a trapezoid, so \[ \frac{m_0}{\rho b} =\ell\eta+\frac12q\eta^2. \] Consequently, \[ \eta =\frac{\sqrt{\ell^2+2q\,m_0/(\rho b)}-\ell}{q} =0.0122822\ \mathrm m. \]

Measure \(u\) along the bucket from its inner vertical wall toward the outer lip. A horizontal strip at height \(y\) above the bottom extends from \(u=0\) to \(u=\ell+qy\). Its centroid therefore gives \[ \bar u =\frac{\displaystyle\frac12\int_0^\eta(\ell+qy)^2\,dy} {\displaystyle\int_0^\eta(\ell+qy)\,dy}. \] Writing \(w=\ell+q\eta\), this becomes \[ \bar u=\frac{\ell^2+\ell w+w^2}{3(\ell+w)} =0.271430\ \mathrm m. \]

At the instant the pestle loses contact with the rice, its supporting reaction vanishes. The two gravitational torques balance: \[ M g r=m_0g(a+\bar u). \] Thus \[ \boxed{r=\frac{m_0}{M}(a+\bar u) =0.0157143\ \mathrm m=1.57\ \mathrm{cm}}. \]

Define the corresponding torque scale \[ K=Mgr=4.6200\ \mathrm{N\,m}. \]

1.2. Beginning and end of spilling

Use coordinates fixed in the bucket:

  • \(s\) runs inward along its upper edge from the outer lip;
  • \(z\) measures depth below that edge.

The sloping end is \(z=s\tan\gamma\), the flat bottom is \(z=h\), and the pivot has coordinates \((R,d)\).

When water reaches the outer lip, its horizontal surface passes through \((0,0)\). In the bucket coordinates its equation is \[ z=s\tan\alpha. \]

The retained water then has a triangular longitudinal cross-section with vertices \[ (0,0),\qquad (hq,h),\qquad (h\cot\alpha,h). \] Its mass is therefore \[ \boxed{m(\alpha)=\frac{\rho b h^2}{2} (\cot\alpha-\cot\gamma) =C(\cot\alpha-q)}. \tag{1} \]

For the \(m_0=1.0\ \mathrm{kg}\) charge, spilling begins when this capacity equals \(m_0\): \[ \cot\alpha_1=q+\frac{m_0}{C}=2.65798. \] Hence \[ \boxed{\alpha_1=20.6176^\circ}. \]

The water disappears when the horizontal free surface becomes parallel to the sloping end: \[ \cot\alpha_2=\cot\gamma, \qquad \boxed{\alpha_2=\gamma=30^\circ}. \]

The triangular construction is consistent with the dimensions: at \(\alpha_1\), its inner endpoint is \(h\cot\alpha_1=0.319\ \mathrm m<L\).

1.3. Balance angle and remaining water mass

The centroid of the overflowing triangular cross-section is \[ s_N=\frac h3(q+\cot\alpha),\qquad z_N=\frac{2h}{3}. \] The dimensions have the useful property \[ z_N=\frac{2h}{3}=0.08\ \mathrm m=d. \] Thus \(TN\) lies along the lever, and the horizontal moment arm of the water is \[ \left[R-\frac h3(q+\cot\alpha)\right]\cos\alpha. \]

The total torque while spilling is \[ \boxed{ \mu(\alpha)=g\cos\alpha \left\{ C(\cot\alpha-q) \left[R-\frac h3(\cot\alpha+q)\right]-Mr \right\}. } \tag{2} \]

Set \(v=\cot\beta-q\). The equilibrium condition becomes \[ Cv\left[R-\frac{2hq}{3}-\frac h3v\right]=Mr. \] The root in the physical range \(\alpha_1<\beta<\alpha_2\) is \[ v=0.560329. \] Therefore \[ \boxed{\beta=\operatorname{arccot}(q+v)=23.5682^\circ}, \] and \[ \boxed{m_1=Cv=0.605156\ \mathrm{kg}}. \]

The distance needed later is \[ \ell_\beta=TN =R-\frac h3(q+\cot\beta) =0.779023\ \mathrm m. \]

2.1. Torque throughout the working cycle

For any water centroid \((s_N,z_N)\), the water’s horizontal moment arm is \[ (R-s_N)\cos\alpha+(d-z_N)\sin\alpha. \] Hence, before spilling, \[ \mu_\uparrow(\alpha) =m_0g\big[(R-s_N)\cos\alpha+(d-z_N)\sin\alpha\big] -K\cos\alpha. \tag{3} \]

The water initially has a trapezoidal cross-section. It becomes triangular when it ceases to touch the inner vertical wall. This occurs at \[ \cot\alpha_c=q+\frac{\rho b\ell^2}{2m_0}, \qquad \alpha_c=2.493^\circ. \]

For \(\alpha_c\leq\alpha\leq\alpha_1\), define \[ \delta(\alpha) =\sqrt{\frac{2m_0}{\rho b(\cot\alpha-q)}}. \] This is the triangle’s height measured perpendicular to the bucket bottom. Its centroid is \[ s_N=hq+\frac{\delta}{3}(\cot\alpha-2q), \qquad z_N=h-\frac{\delta}{3}. \] These expressions, together with (3), determine the rising branch before spilling. Between \(\alpha_1\) and \(\alpha_2\), use (2).

Once empty, the bucket contributes no torque: \[ \boxed{\mu_{\mathrm{empty}}(\alpha)=-K\cos\alpha}. \]

The graph can be drawn from the following values:

\(\alpha\) Rising-stroke torque \(\mathrm{(N\,m)}\) Empty return torque \(\mathrm{(N\,m)}\)
\(0^\circ\) \(0\) at lift-off \(-4.620\)
\(2.493^\circ\) \(+0.944\) \(-4.616\)
\(5^\circ\) \(+1.561\) \(-4.602\)
\(10^\circ\) \(+2.068\) \(-4.550\)
\(15^\circ\) \(+2.367\) \(-4.463\)
\(\alpha_1=20.618^\circ\) \(+2.687\) \(-4.324\)
\(\beta=23.568^\circ\) \(0\) \(-4.235\)
\(\alpha_2=30^\circ\) \(-4.001\) \(-4.001\)

The complete cycle has these features:

  1. At \(\alpha=0\), filling raises the gravitational torque from \(-K\) to zero. This is a vertical segment in the torque–angle plot. The mortar’s reaction keeps the lever stationary during this stage.
  2. On lifting, the torque rises from zero to its maximum \(+2.687\ \mathrm{N\,m}\) at \(\alpha_1\).
  3. Spilling makes the torque decrease, cross zero at \(\beta\), and reach \(-4.001\ \mathrm{N\,m}\) at \(\alpha_2\).
  4. From \(\alpha_2\) to the maximum angle \(\alpha_0\), and throughout the empty return, the torque follows \(-K\cos\alpha\). These outward and return curves coincide for \(\alpha\geq\alpha_2\).

In particular, \(\mu(0)\) has different values at different stages: zero at lift-off and \(-4.620\ \mathrm{N\,m}\) with the empty bucket.

2.2. Work and its geometric interpretation

An infinitesimal rotational work is \[ dW=\mu\,d\alpha, \] with angles measured in radians.

The total work over a cycle is \[ W_{\mathrm{total}} =\oint\mu\,d\alpha =\int_0^{\alpha_0} \big[\mu_\uparrow(\alpha)-\mu_\downarrow(\alpha)\big]\,d\alpha. \] The branches coincide above \(\alpha_2\), so \[ \boxed{ W_{\mathrm{total}} =\int_0^{\alpha_2} \big[\mu_\uparrow(\alpha)+K\cos\alpha\big]\,d\alpha. } \] Geometrically, this is the area enclosed between the rising branch, the empty return branch, and the filling segment at \(\alpha=0\).

The empty lever starts its return from rest at \(\alpha_0\). Its gravitational potential-energy loss becomes the kinetic energy delivered at impact: \[ \boxed{ W_{\mathrm{pounding}} =\int_{\alpha_0}^{0}(-K\cos\alpha)\,d\alpha =K\sin\alpha_0. } \] It is the positive area between the empty-return curve and the angle axis, from \(0\) to \(\alpha_0\).

If escaping water carries kinetic energy away, that energy reduces the fraction of the cycle’s work delivered to the rice. Under the approximation in 2.3 that this energy is negligible, \[ \boxed{W_{\mathrm{total}}=W_{\mathrm{pounding}}}. \] Equivalently, the positive and negative signed areas on the complete rising stroke cancel, since that stroke starts and ends at rest: \[ \int_0^{\alpha_0}\mu_\uparrow(\alpha)\,d\alpha=0. \]

2.3. Maximum angle and pounding work

Let \[ E_2=\int_0^{\alpha_2}\mu_\uparrow(\alpha)\,d\alpha. \] This is the kinetic energy remaining when the last water leaves.

Connecting the tabulated rising-branch points by straight segments gives \[ E_2\approx0.520\ \mathrm J. \] From \(\alpha_2\) to \(\alpha_0\), the empty lever converts this energy into gravitational potential energy: \[ E_2=K(\sin\alpha_0-\sin\alpha_2). \] Thus the polygonal estimate is \[ \sin\alpha_0 =\frac12+\frac{0.520}{4.620}, \] giving \[ \alpha_0\approx37.8^\circ, \qquad W_{\mathrm{pounding}} =K\sin\alpha_0\approx2.83\ \mathrm J. \]

Integrating the curved torque branches more accurately gives \[ \int_0^{\alpha_1}\mu_\uparrow\,d\alpha=0.6796\ \mathrm J, \qquad \int_{\alpha_1}^{\alpha_2}\mu_\uparrow\,d\alpha=-0.1779\ \mathrm J. \] Hence \(E_2=0.5016\ \mathrm J\), yielding \[ \boxed{\alpha_0\approx37.5^\circ,\qquad W_{\mathrm{pounding}}\approx2.8\ \mathrm J}. \]

3.1.1. Equilibrium with continuous overflow

Continuous overflow makes the water mass a function of angle given by (1), on both directions of motion.

Near \(\beta\), increasing \(\alpha\) reduces the water mass and produces a negative torque. Decreasing \(\alpha\) increases the retained water mass and produces a positive torque.

Thus the graph passes through \((\beta,0)\) with negative slope, and \[ \boxed{\alpha=\beta\text{ is a stable equilibrium}.} \] In the stated idealization, small disturbances produce undamped oscillations.

3.1.2. Linearized torque

Write (2) as \[ \mu(\alpha)=g\cos\alpha\,F(\cot\alpha), \] where \[ F(c)=C(c-q)\left[R-\frac h3(c+q)\right]-Mr. \] At equilibrium \(F(\cot\beta)=0\). Also, \[ \frac{dF}{dc}=C\left(R-\frac{2h}{3}c\right), \qquad \frac{d(\cot\alpha)}{d\alpha}=-\csc^2\alpha. \]

Therefore \[ \mu'(\beta) =-\frac{gC\cos\beta}{\sin^2\beta} \left(R-\frac{2h}{3}\cot\beta\right). \] Define \[ \boxed{ \kappa= \frac{gC\cos\beta}{\sin^2\beta} \left(R-\frac{2h}{3}\cot\beta\right) =45.91\ \mathrm{N\,m\,rad^{-1}}. } \] For a small angular displacement in radians, \[ \boxed{\mu(\beta+\Delta\alpha) =-\kappa\,\Delta\alpha+O((\Delta\alpha)^2)}. \]

3.1.3. Equation of motion and period

To first order, evaluate the water’s inertia at equilibrium. Approximating its small contribution by a mass at its centroid gives \[ J_\beta\approx I+m_1\ell_\beta^2 =12+0.6052(0.7790)^2 \approx12.37\ \mathrm{kg\,m^2}. \] The water’s moment about its own centroid is only about \(0.0026\ \mathrm{kg\,m^2}\), so including its spatial extent has negligible effect on the quoted period.

Changes of inertia multiply small accelerations or velocities and contribute beyond the linear approximation. Thus, putting \(x=\alpha-\beta\), \[ \boxed{J_\beta\ddot x+\kappa x=0}. \] This is the harmonic-oscillator equation, with \[ \omega=\sqrt{\frac{\kappa}{J_\beta}} \approx1.927\ \mathrm{s^{-1}}. \]

For \(x(0)=\Delta\alpha\) and \(\dot x(0)=0\), \[ \boxed{\alpha(t)=\beta+\Delta\alpha\cos(\omega t)}. \] The period is \[ \boxed{ \tau=2\pi\sqrt{\frac{J_\beta}{\kappa}} \approx3.26\ \mathrm s. } \]

3.2. Flow required for amplitude \(1^\circ\)

The overflowing bucket’s capacity obeys \[ \frac{dm}{d\alpha}=-C\csc^2\alpha. \] During the downward part of the oscillation, this capacity increases. To maintain overflow, the incoming mass flow must satisfy \[ \Phi\geq\frac{dm}{dt} =-C\csc^2\alpha\,\dot\alpha. \]

For small oscillations of amplitude \(A\), \[ \alpha=\beta+A\cos\omega t. \] To first order, \[ \frac{dm}{dt} =\frac{C}{\sin^2\beta}A\omega\sin\omega t. \] Its maximum occurs at the downward passage through equilibrium. Therefore \[ \boxed{\Phi_{\min}(A) =\frac{C}{\sin^2\beta}\,\omega A}. \tag{4} \]

Numerically, \[ \frac{C}{\sin^2\beta}=6.755\ \mathrm{kg\,rad^{-1}}. \] For \(A=1^\circ=\pi/180\) radians, \[ \Phi_1 =6.755(1.927)\frac{\pi}{180} =0.227\ \mathrm{kg\,s^{-1}}. \] Hence \[ \boxed{\Phi_1\approx0.23\ \mathrm{kg\,s^{-1}}}. \]

3.3. Estimated flow threshold for cessation of pounding

Use the marginal raised oscillation whose lower turning point is \(\alpha_1\). Its centre is the equilibrium angle \(\beta\), so its amplitude is \[ \boxed{A_2=\beta-\alpha_1 =2.9506^\circ=0.051497\ \mathrm{rad}}. \] This limiting oscillation remains above the horizontal position, so the pestle does not reach the rice.

Applying the same overflow condition (4), \[ \Phi_2 \approx\frac{C}{\sin^2\beta}\omega(\beta-\alpha_1). \] Equivalently, \[ \Phi_2\approx \Phi_1\frac{\beta-\alpha_1}{1^\circ}. \] Thus \[ \Phi_2\approx0.227(2.9506) =0.670\ \mathrm{kg\,s^{-1}}. \]

The finite-amplitude correction can be checked while retaining a harmonic trajectory. On its descending branch, \[ -\dot\alpha =\omega\sqrt{A_2^2-(\alpha-\beta)^2}, \] so the exact capacity-growth rate is \[ \frac{dm}{dt} =\frac{C\omega}{\sin^2\alpha} \sqrt{A_2^2-(\alpha-\beta)^2}. \] Its maximum is approximately \(0.69\ \mathrm{kg\,s^{-1}}\), consistent with the estimate \[ \boxed{\Phi_2\approx0.7\ \mathrm{kg\,s^{-1}}}. \]

The solution was checked against the official solution.