A mass of 40 kg is rotated by a string at a constant speed, V, in a vertical circle of 1m. If the minimum tension is 3N, Calculate, (i) Velocity (ii) Tmax (iii) Tension when the string is just horizontal

The minimum tension occurs at the top of the vertical circle, where $\theta = 180^\circ$. At this point, $\cos \theta = \cos 180^\circ = -1$.

Substitute $T_{min} = 3 \, \text{N}$ and $\cos \theta = -1$ into the tension formula:

The velocity is $\boxed{3.14 \, \text{m/s}}$.

The maximum tension occurs at the bottom of the vertical circle, where $\theta = 0^\circ$. At this point, $\cos \theta = \cos 0^\circ = 1$.

Substitute $V^2 = 9.875 \, \text{m}^2/\text{s}^2$ and $\cos \theta = 1$ into the tension formula:

The maximum tension is $\boxed{787 \, \text{N}}$.

When the string is horizontal, the angle $\theta = 90^\circ$ (or $270^\circ$). At this point, $\cos \theta = \cos 90^\circ = 0$.

Substitute $V^2 = 9.875 \, \text{m}^2/\text{s}^2$ and $\cos \theta = 0$ into the tension formula:

The tension when the string is just horizontal is $\boxed{395 \, \text{N}}$.