Q.A tiny stone of mass 20 g is tied to a practically massless, inextensible, flexible string and whirled along vertical circles. Speed of the stone is 8 m/s when the centripetal force is exactly equal to the force due to the tension. Calculate minimum and maximum kinetic energies of the stone during the entire circle. Determine the angular position of the string when the force due to tension is numerically equal to weight of the stone. Use g = 10 m/s² and length of the string = 1.8 m

Tension equals the full centripetal force only where the string is horizontal, so the 8 m/s speed belongs to the horizontal position; energy conservation then fixes the extremes. …

A stone of mass 20 g on a string of length 1.8 m is whirled in vertical circles; its speed is 8 m/s at the instant the string is exactly horizontal (i.e. at position C or D, where the centripetal force equals the tension alone, since weight is tangential there) -- use g = 10 m/s^2. Since the vertical displacement between the horizontal-string position and the top/bottom is exactly r each way, energy conservation from that 8 m/s reference speed gives the kinetic energies at the top (minimum, since it is highest) and bottom (maximum, since it is lowest): $KE_{min}=\frac{1}{2}m(8)^2-mgr$ and $KE_{max}=\frac{1}{2}m(8)^2+mgr$, evaluating to 0.28 J and 1.0 J respectively (giving $v_{max}=10$ m/s at the bottom). The second part sets up the angular position $\theta$ (measured from the LOWERMOST point, where $\theta=0$) at which the string te …

Use energy bookkeeping from the horizontal position directly: $K.E._{top} = \frac{1}{2}mv_H^2 - mgr$ and $K.E._{bottom} = \frac{1}{2}mv_H^2 + mgr$ with $v_H=8$ m/s, $r=1. …

Assuming 8 m/s is the speed at the TOP or BOTTOM — the equality of tension and centripetal force holds only where the weight is entirely tangential (string horizont …