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(B) Let the mass of the particle be $m$,the radius of the circle be $r$,and the angular velocity be $\omega$. The centripetal force is provided by the

Vedclass · Accepted Answer

$T_1 > T_2 > T_3, T_2 = T_4$

Vedclass · Answer

(B) Let the mass of the particle be $m$,the radius of the circle be $r$,and the angular velocity be $\omega$. The centripetal force is provided by the net radial force.At point $A$ (bottom): The tension $T_1$ acts upwards and weight $mg$ acts downwards. The net force is $T_1 - mg = m\omega^2 r$. Thus,$T_1 = mg + m\omega^2 r$ ... $(1)$At point $B$ (right side): The tension $T_2$ acts towards the center. The weight $mg$ acts downwards,which is perpendicular to the radius. Thus,$T_2 = m\omega^2 r$ ... $(2)$At point $C$ (top): Both tension $T_3$ and weight $mg$ act downwards. The net force is $T_3 + mg = m\omega^2 r$. Thus,$T_3 = m\omega^2 r - mg$ ... $(3)$At point $D$ (left side): The tension $T_4$ acts towards the center. The weight $mg$ acts downwards,which is perpendicular to the radius. Thus,$T_4 = m\omega^2 r$ ... $(4)$Comparing the equations: $T_1 = mg + m\omega^2 r$,$T_2 = m\omega^2 r$,$T_3 = m\omega^2 r - mg$,and $T_4 = m\omega^2 r$.Clearly,$T_1 > T_2 = T_4 > T_3$. Therefore,$T_1 > T_2 > T_3$ and $T_2 = T_4$.

$A$ particle tied to one end of a string is being rotated in a vertical circle with constant frequency. The tension in the string at points $A, B, C$ and $D$ are $T_1, T_2, T_3$ and $T_4$ respectively. Then:

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