(A) In a conical pendulum,the forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$.Resolving the tension $T$ in

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(A) In a conical pendulum,the forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$.Resolving the tension $T$ into components:$T \cos \theta = mg$ (vertical equilibrium)$T \sin \theta = F_c$ (where $F_c$ is the centripetal force)From the geometry of the pendulum,$\sin \theta = \frac{r}{L}$.Then,$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{r^2}{L^2}} = \frac{\sqrt{L^2-r^2}}{L}$.From the first equation,$T = \frac{mg}{\cos \theta} = \frac{mgL}{\sqrt{L^2-r^2}}$.Substituting $T$ into the centripetal force equation:$F_c = T \sin \theta = \left( \frac{mgL}{\sqrt{L^2-r^2}} \right) \times \left( \frac{r}{L} \right) = \frac{mgr}{\sqrt{L^2-r^2}}$.

Class 11 Physics 2 Motion in a Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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In a conical pendulum,the bob of mass $m$ moves in a horizontal circle of radius $r$ with uniform speed $V$. The string of length $L$ describes a cone of semi-vertical angle $\theta$. The centripetal force acting on the bob is $(g = \text{acceleration due to gravity})$

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A
$\frac{mgr}{\sqrt{L^2-r^2}}$
B
$\frac{mgr}{L^2-r^2}$
C
$\frac{\sqrt{L^2-r^2}}{mgL}$
D
$\frac{mgL}{\sqrt{L^2-r^2}}$

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2 Motion in a Plane

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Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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In a conical pendulum,the bob of mass $m$ moves in a horizontal circle of radius $r$ with uniform speed $V$. The string of length $L$ describes a cone of semi-vertical angle $\theta$. The centripetal force acting on the bob is $(g = \text{acceleration due to gravity})$

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