What would be the acceleration of a ball moving in uniform circular motion? Question in a book:

A ball of $mass = 0.5kg$ is attached to the end of an inelastic string
  of length $L=0.5m$. The other end of the string is fixed. The ball is
  made to move on a horizontal circular path about the vertical axis
  through the fixed end of the string. The maximum tension that the
  string can bear is 324 N. The maximum possible angular velocity of the
  ball is 36 rad/s, the acceleration of the ball along the string is?.

Solution given: 

Since the string is inelastic acceleration along the string is zero.

My problem with the solution:
Why on the earth would it be zero. If the ball is rotating with $\omega$ then shouldn't the centripetal acceleration which is along the string directed towards the center be $r\omega^2$? But it isn't mentioned in the problem with what $\omega$ the string is rotating. Then how to go about solving the problem?
PS: I am a physics tutor for 11th and 12th grades.
 A: The question is very poorly worded and confusing.
The fact that the ball is forced to move in a circular path means that the radial velocity is zero but the radial acceleration is $a_r=r\omega^2$ as you correctly point out. Sufficient information is provided to calculate this acceleration, assuming that the angular velocity $\omega$ and tension $T$ are the maximum possible.
The fact that the string is inextensible is irrelevant, because a ball at the end of an elastic string can also be forced to move in a perfect circle.
What is described in the question is a conical pendulum. The string does not lie along the radius of the circle which the ball moves in. The radius of this circle is $r=L\sin\theta$ where $\theta$ is the inclination of the string to the vertical. The tension $T$ in the string is such that
$T\cos\theta=mg$
$T\sin\theta=ma_r$
where $a_r$ is the centripetal acceleration. From these you can eliminate $\theta$ to get
$a_r^2=(\frac{T}{m})^2-g^2$.  
There is a non-zero component of centripetal acceleration along the direction of the string - ie $a_r\sin\theta$.
A: In polar coordinates: $P(r,\theta)$
As the string is inelastic:
$$r=const. \quad \to \quad \dot{r}=\ddot{r}=0 $$
$$ \dot{\theta}=const.=36\frac{rad}{s} \quad \to \quad \ddot{\theta}=0 $$
For curvilinear the acceleration towards the center $a_r$ is given by:
$$ a_r=\ddot{r}-r\cdot{\dot{\theta}}^2 $$
$$ \to a_r=-0.5m\cdot\left(36\frac{rad}{s}\right)^2=-648\frac{m}{s^2} $$
