= Solution
Let $\Delta\rho=\rho_p-\rho_f>0$. A spherical grain has <submerged weight>
$$
G=\frac\pi6\Delta\rho\,gd^3.
$$
Neglect lift and contact torque, and use a sliding <static friction> model with normal reaction $R=G$. The inertial <quadratic drag> is
$$
F_D=\frac12C_D\rho_fU^2\frac{\pi d^2}{4}=\frac{C_D\pi d^2}{8}\tau,
$$
where the last equality fixes the convention $\tau=\rho_fU^2$. Equivalently $U$ is the bed <shear velocity> and $C_D$ is an effective <drag coefficient> referred to it. A literal grain-level flow speed can differ from the <shear velocity>; that conversion must then be absorbed into $C_D$.
Downstream sliding begins when $F_D>\mu_sR$. Defining the <Shields parameter> by $\Theta=\tau/(\Delta\rho\,gd)$, the threshold <force balance> is
$$
\frac{C_D\pi d^2}{8}\tau_{\mathrm{th},0}=\mu_s\frac\pi6\Delta\rho\,gd^3,
\qquad
\boxed{\Theta_{\mathrm{th},0}=\frac{4\mu_s}{3C_D}}.
$$
\b[The grain moves downstream above this threshold in the stated sliding model.] Real grain motion can instead involve lift, rolling, irregular contacts, or viscous drag; the printed constant belongs to the particular inertial-drag convention and <force balance> above.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-345-grain.png]
{title=Grain force balances on horizontal and inclined beds}
The normal contact <force> and resisting <static friction> balance the <drag force> and <submerged weight> at impending motion. The right panel uses locally bed-tangent <drag force>.
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