= Solution
An <impulse> is $\mathbf J=\int\mathbf F\,dt$, giving $\Delta\mathbf P=\mathbf J$. For a short localized <impulse> at <displacement> $\mathbf r$ from the <centre of mass>, the <angular impulse> is $\Delta\mathbf L_{\rm CM}=\mathbf r\times\mathbf J$; for a changing lever arm use its instantaneous value inside the <integral>. For a uniform solid sphere with <volume> <density> $\rho_V$, direct integration about the $z$ axis gives
$$
I=\rho_V\int_0^a r^4\,dr\int_0^\pi\sin^3\theta\,d\theta\int_0^{2\pi}d\phi=\frac{8\pi}{15}\rho_Va^5=\boxed{\frac25Ma^2}.
$$
Take forward translation $u$ and forward-rolling clockwise spin $\Omega$ as positive. Neglect the table's tangential <impulse> during the much shorter cue impact. Immediately afterwards,
$$
\boxed{u_0=\frac{I_p}{M},\qquad\Omega_0=\frac{dI_p}{I}=\frac{5dI_p}{2Ma^2}.}
$$
The bottom contact point has slip <velocity> $s_0=u_0-a\Omega_0$. The <cue height for immediate rolling> is therefore \b[$d=2a/5$].
For the <impulse-driven slipping-to-rolling transition of a sphere> at an arbitrary admissible impact height, let $J_f$ be the signed <friction> <impulse> accumulated before slipping ends. Translation and spin then obey $u_f=u_0+J_f/M$ and $\Omega_f=\Omega_0-aJ_f/I$. Impose $u_f=a\Omega_f$ and eliminate $J_f$:
$$
\boxed{u_f=\frac{u_0+(I/(Ma))\Omega_0}{1+I/(Ma^2)}=\frac{5I_p}{7M}\left(1+\frac da\right).}
$$
This expression handles both sliding directions. If $d<2a/5$, the contact point slips forward and <friction> slows the centre; if $d>2a/5$, excess spin makes it slip backward and <friction> accelerates the centre. <friction> still dissipates total <mechanical energy>. For $\mu_d>0$, $\dot s=-(7/2)\mu_dg\,\operatorname{sgn}s$, so the slipping time is $2|s_0|/(7\mu_dg)$. Its duration depends on <friction>, while the final <speed> does not. With zero <friction> no slipping-to-rolling transition occurs unless rolling was present from the outset.
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