Consider the set: $$\tilde\Omega=\{(t,s)\in [0,\infty)\times [0,l): t\leq \rho(s)\}$$ where we have set $\rho(s)=r(\gamma(s))$. The starlike property implies that the map $
\Phi:\tilde\Omega\to \Omega
$ defined by $$\Phi(t,s)=\gamma(s)+tN(s)$$ is a diffeomorphism. Let us compute the Euclidean metric tensor in the coordinates $(t,s)$. Write $\gamma'(s)=T(s)$ for the unit tangent vector to $\gamma$ and observe that $N'(s)=k(s)T(s)$, where $k(s)$ is the curvature of $\Sigma_1$ which is everywhere non-negative because $\Sigma_1$ is convex. Then: $$\twosystem
{d\Phi(\dfrac{{\partial}}{{\partial}t})=N(s)}
{d\Phi(\dfrac{{\partial}}{{\partial}s})=(1+tk(s))T(s)}$$ If we set $\Theta(t,s)=1+t k(s)$ the metric tensor is: $$g={{\begin{pmatrix}}{1&0\\}{0&\Theta^2\\}{\end{pmatrix}}}$$ and an orthonormal basis is then $(e_1,e_2)$, where $$e_1=\dfrac{{\partial}}{{\partial}t}, \quad e_2=\dfrac{1}{\Theta}\dfrac{{\partial}}{{\partial}s}.$$ In these coordinates, our function $\psi$ is written: $$\psi(t,s)=\dfrac{t}{\rho(s)}.$$ Now $$\twosystem
{{\langle{\nabla\psi},{e_1}\rangle}={\dfrac{{\partial}\psi}{\bdt}}=\dfrac{1}{\rho(s)}}
{{\langle{\nabla\psi},{e_2}\rangle}=\dfrac{1}{\Theta}{\dfrac{{\partial}\psi}{\bds}}=-\dfrac{t\rho'(s)}{\Theta(t,s)\rho(s)^2}}.$$ It follows that $${\lvert{\nabla\psi}\rvert}^2=\dfrac{1}{\rho^2}+\dfrac{t^2\rho'^2}{\Theta^2\rho^4}=
\dfrac{\Theta^2\rho^2+t^2\rho'^2}{\Theta^2\rho^4}.$$ Recall the radial gradient, which is the orthogonal projection of $\nabla\psi$ on the ray, whose direction is given by $e_1$. If we fix $x\in\Sigma_1$, we have $$\theta_x(t)=\text{angle between $\nabla\psi$ and $e_1$}$$ and we have to study the function $$h(t)=\cos\theta_x(t)=\dfrac{{\langle{\nabla\psi},{e_1}\rangle}}{{\lvert{\nabla\psi}\rvert}}=\dfrac{1}{\rho(s){\lvert{\nabla\psi}\rvert}}$$ for a fixed $s$. From the above expression of ${\lvert{\nabla\psi}\rvert}$ and a suitable manipulation we see $$h(t)^2=\dfrac{\Theta^2}{\Theta^2+t^2g^2}$$ where $g=\rho'(s)/\rho(s)$. Now $$\begin{aligned}
\dfrac{d}{dt}\dfrac{\Theta^2}{\Theta^2+t^2g^2}&=\dfrac{2t\Theta g^2}{(\Theta^2+t^2g^2)^2}
(t{\dfrac{{\partial}\Theta}{\bdt}}-\Theta)\\
\end{aligned}$$ As $\Theta(t,s)=1+tk(s)$ one sees that $t{\dfrac{{\partial}\Theta}{\bdt}}-\Theta=-1$ hence $$\dfrac{d}{dt}h(t)^2=-\dfrac{2t\Theta g^2}{(\Theta^2+t^2g^2)^2}\leq 0$$ Hence $h(t)^2$ is non-increasing and, as $h(t)$ is positive, it is itself non-increasing.

[**Proof 