Q.Using energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object.

Using energy conservation between the top/bottom and force equations at each of the 4 key points gives $v_{A,min}=\sqrt{rg}$, $v_{B,min}=\sqrt{5rg}$, $v_{C,min}=v_{D,min}=\sqrt{3rg}$; zero speed at the top is possible ONLY for a rigid rod (never a string, since a string cannot push); and $T_B-T_A=6mg$, independent of v and r.

Consider a bob (string case) undergoing vertical circular motion under gravity, radius r (the string length). At the UPPERMOST point A, both mg and the string tension $T_A$ point down (towards the centre): $mg+T_A=\frac{mv_A^2}{r}$. Since a string can only pull (never push), $T_A\geq0$ always; the borderline, minimum-energy case is $T_A=0$, giving $v_{A,min}=\sqrt{rg}$ --- Eq. (1.10). At the LOWERMOST point B, $T_B-mg=\frac{mv_B^2}{r}$. Falling from A to B, the bob drops through vertical height 2r under gravity alone, so energy conservation gives $\frac{1}{2}mv_B^2=\frac{1}{2}mv_A^2+mg(2r)$, i.e. $v_B^2=v_A^2+4gr$; substituting $v_{A,min}^2=rg$ gives $v_{B,min}^2=rg+4gr=5rg$, so $v_{B,min}=\sqrt{5rg}$ --- Eq. (1.13). At the two HORIZONTAL positions C and D, the tension alone (weight being tangential there) supplies the centripetal force; working through the corresponding energy and force equations (dropping through height r from A, or rising through height r from B) gives $v_{C,min}=v_{D,min}=\sqrt{3rg}$.

Zero speed at the uppermost point is possible ONLY in the alternative case where the bob is tied to a RIGID ROD instead of a string (section 1.4.3's Case II). A rod, unlike a string, can PUSH as well as pull, so it does not need any minimum tension (and hence no minimum speed) to keep the bob on the circular path at the top -- the rod can simply push the bob inward even at (practically) zero speed there. For a string, by contrast, zero speed at the top is never possible, since that would require $T_A<0$ (the string 'pushing', which it cannot do), so the string would instead go slack and the bob would fall away from the circular path.

Finally, subtracting the force equation at A from the force equation at B: $(T_B-mg)-(-mg-T_A) \;\Rightarrow\; T_B-T_A-2mg=\frac{m}{r}(v_B^2-v_A^2)=\frac{m}{r}(4gr)=4mg$ so $T_B-T_A=4mg+2mg=6mg$ Every term involving r or the actual speeds has cancelled out in this derivation (the $4gr$ from energy conservation exactly cancels the $1/r$ from the force equations), leaving a result that depends ONLY on the weight mg of the object -- proving that the extreme tension difference is fixed at $6mg$ for ANY valid vertical circular motion under gravity, not merely the minimum-energy case used to derive it.

At each key point, write the force equation (resultant of weight and tension/normal force = centripetal force) and combine it with energy conservation between points to eliminate the unknown speeds; set the string tension to its minimum allowed value (zero) to get the minimum-speed conditions.

Assuming the string case also permits zero speed at the top (confusing it with the rigid-rod case), or forgetting that r cancels out of the final tension-difference result, and instead trying to compute $T_B-T_A$ separately for different radii expecting different answers.