[Physics] Tension at extreme of a rotating rope of mass $m$

Question

A uniform rope of mass per unit length $\lambda$, length $\ell$ is attached to shaft that is rotating at constant
angular velocity $\omega$. Find the tension in the rope as a function of distance from the shaft.
You may ignore the effect of gravitation.

In its solution, we can consider any general differential element $dx$ at a distance $x$ from the axis of rotation, and obtain the equation
$$ -dT=\lambda \omega^2 x dx $$
Once, we integrate it, we does it from $x$ to $\ell$ and take the corresponding tension from $T(x)$ to $0$.
But can anyone tell why $T(\ell) = 0$, cause without tension at the extreme end what can provide the the mass $dm$(at the extreme) the required centripetal force?
By the way , final answer is
$$T(x)=\frac{\lambda \omega^2(l^2-x^2)}{2}$$

Answer 1

You can also consider the problem in a global way by considering the piece of rope included between $x$ and $l$.

Its mass is $m(x)=\lambda (l-x)$ and its center of gravity is in ${{x}_{G}}=x+\frac{1}{2}(l-x)=\frac{1}{2}(l+x)$.

Its acceleration is $-m(x){{\omega }^{2}}{{x}_{G}}=-\lambda (l-x){{\omega }^{2}}\frac{1}{2}(l+x)=-\frac{1}{2}\lambda {{\omega }^{2}}({{l}^{2}}-{{x}^{2}})$

The tension of the rope is necessary to maintain this acceleration $T(x)=-\frac{1}{2}\lambda {{\omega }^{2}}({{l}^{2}}-{{x}^{2}})$

Maybe this way to consider the problem may help you to better understand why the tension is $0$ at the extremity or the rope : no mass to accelerate.

Sorry for my poor english !