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Question

(C) The tension in a string for a particle moving in a vertical circle at an angle $	heta$ from the lowest point is given by $T = \frac{mv^2}{r} + mg

Vedclass · Accepted Answer

$T_1 > T_2$

Vedclass · Answer

(C) The tension in a string for a particle moving in a vertical circle at an angle $	heta$ from the lowest point is given by $T = \frac{mv^2}{r} + mg \cos 	heta$.From the principle of conservation of energy, the velocity $v$ at an angle $	heta$ is related to the velocity at the lowest point $u$ by $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mgr(1 - \cos 	heta)$.Substituting $v^2$ into the tension equation, we get $T = \frac{m}{r}(u^2 - 2gr(1 - \cos 	heta)) + mg \cos 	heta = \frac{mu^2}{r} - 2mg + 3mg \cos 	heta$.Since $T$ is a function of $\cos 	heta$, and $\cos 30^\circ > \cos 60^\circ$, it follows that $T_1 > T_2$.

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 $= 4 mg$

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

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