Newtonian Mechanics – Understanding Instant Axis of Rotation

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The given question is:

A man is rotating a stone of mass 10 kg tied at the end of a light rope in a circle of radius 1m. To
do this, he continuously moves his hand in a circle of radius 0.6 m. Assume, both circular motions
to be occurring in the same horizontal plane. What is the maximum speed with which he can throw
the stone, if he can exert a pull not exceeding 1250 N on the string?

Now I am lacking some conceptual clarity in this area, So could someone explain the logic behind this question so that I can also get the conceptual clarity I need in this area? The answer given is 10m/s.

Best Answer

If the stone is moving in a circle - as we are told - then the net horizontal force on the stone must be directed towards the centre of the circle. And the only horizontal force on the stone is the tension in the rope. So the centre of the stone's circle must lie somewhere along the line of the rope i.e. along the line between the stone and the man's hand (including its extension).

So if we call the centre of the circle O, the position of the man's hand A and the position of the stone B then O, A and B always lie on a straight line. We know that the distance OA is $0.6$ m and OB is $1$ m. The order is either OAB - in which case the rope has a length of $0.4$m - or AOB - in which case the rope has a length of $1.6$ m.

In either case the tension in the rope is $T=\frac {mv^2}{r}$ where $m=10$ kg, $r=1$ m and $v$ is the speed of the stone.

(Note that this means the given answer of $v=10$ m/s is incorrect)

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