No it is still zero.

Take $\theta$ to be the angle between the point at the bottom of the cylinder, C and P. Take the radius of the cylinder to be $R$ and max length of rope $L$.

The vector from C to P is $$r_{CP} = R(\sin \theta,-\cos\theta).$$

The vector from P to the particle is $$r_{T}=(L-R\theta)(\cos\theta,\sin\theta).$$

So the position vector of the particle is $r = r_{CP}+r_{T}.$ The tangent vector to the path of the particle is $$\frac{dr}{d\theta}=-(L-R\theta)(\cos\theta,\sin\theta)$$ which is perpendicular to the direction of tension $r_T$,

No it is still zero.

Take $\theta$ to be the angle between the point at the bottom of the cylinder, C and P. Take the radius of the cylinder to be $R$ and max length of rope $L$.

The vector from C to P is $$r_{CP} = R(\sin \theta,-\cos\theta).$$

The vector from P to the particle is $$r_{T}=(L-R\theta)(\cos\theta,\sin\theta).$$

So the position vector of the particle is $r = r_{CP}+r_{T}.$ The tangent vector to the path of the particle is $$\frac{dr}{d\theta}=(L-R\theta)(-\sin\theta,\cos\theta)$$ which is perpendicular to the direction of tension $r_T$,