In vertical circular motion of a bob,match the entries of List-$I$ with entries of List-$II$. Here,$v_0$ is the velocity of the bob at the lowest point.
List-$I$ (Speed at lowest point) List-$II$ (Possible situation) $(P) v_0 = \sqrt{5 g \ell}$ $(1)$ Tension at lowest point $= 6 mg$ $(Q) v_0 = \sqrt{g \ell}$ $(2)$ String will slack for a finite time $(R) v_0 = 2 \sqrt{g \ell}$ $(3)$ Bob will oscillate $(S) v_0 = 3 \sqrt{g \ell}$ $(4)$ Tension at highest point $= 4 mg$
| List-$I$ (Speed at lowest point) | List-$II$ (Possible situation) |
| $(P) v_0 = \sqrt{5 g \ell}$ | $(1)$ Tension at lowest point $= 6 mg$ |
| $(Q) v_0 = \sqrt{g \ell}$ | $(2)$ String will slack for a finite time |
| $(R) v_0 = 2 \sqrt{g \ell}$ | $(3)$ Bob will oscillate |
| $(S) v_0 = 3 \sqrt{g \ell}$ | $(4)$ Tension at highest point $= 4 mg$ |
- A$P-1, Q-2, R-3, S-4$
- B$P-4, Q-2, R-3, S-1$
- C$P-1, Q-3, R-2, S-4$
- D$P-4, Q-3, R-2, S-1$
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$A$ stone of mass $m$ tied to the end of a string revolves in a vertical circle of radius $R$. The net forces at the lowest and highest points of the circle directed vertically downwards are:
Lowest Point Highest Point $(a) \ mg - T_1$ $mg + T_2$ $(b) \ mg + T_1$ $mg - T_2$ $(c) \ mg + T_1 - \frac{mv_1^2}{R}$ $mg - T_2 + \frac{mv_2^2}{R}$ $(d) \ mg - T_1 - \frac{mv_1^2}{R}$ $mg + T_2 + \frac{mv_2^2}{R}$
$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.
| Lowest Point | Highest Point |
| $(a) \ mg - T_1$ | $mg + T_2$ |
| $(b) \ mg + T_1$ | $mg - T_2$ |
| $(c) \ mg + T_1 - \frac{mv_1^2}{R}$ | $mg - T_2 + \frac{mv_2^2}{R}$ |
| $(d) \ mg - T_1 - \frac{mv_1^2}{R}$ | $mg + T_2 + \frac{mv_2^2}{R}$ |
$T_1$ and $v_1$ denote the tension and speed at the lowest point. $T_2$ and $v_2$ denote corresponding values at the highest point.
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View SolutionIn a vertical circle of radius $r$,at what point in its path does a particle have tension equal to zero if it is just able to complete the vertical circle?
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