A particle of mass m is suspended from point O and undergoes circular motion in horizontal plane as conical pendulum as shown in figure.

Engineering

Physics

Horizontal Circular Motion

Centripetal and Centrifugal Force

Circular Motion in Vertical Plane

Question

A particle of mass m is suspended from point O and undergoes circular motion in horizontal plane as conical pendulum as shown in figure.

Angular momentum of particle about centre of circle remains constant.

Angular momentum of particle about point of suspension does not remains constant.

Average torque about axis OC during half rotation is zero

Average force during half rotation is $\frac{2 mgtanθ}{π}$

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(A, B & D)

τ_av × Δt = L₂ – L₁ = ΔL
${\vec{L}}_{1} = - Lcosθ \hat{i} + Lsinθ \hat{j}$ ,   ${\vec{L}}_{2} = Lcosθ \hat{i} + Lsinθ \hat{j}$
${\vec{L}}_{2} - {\vec{L}}_{1}$ = $(2 Lcosθ \hat{i})$
magnitude is same but direction of angular momentum is continuously changing. So $\vec{L}$ is not constant
(B) about centre is constant
(C) Tcosθ = mg       ...(1)
Tsinθ = mω²ℓsinθ   ...(2)
mω²ℓcosθ = mg
ω² = $\frac{g}{l cosθ}$ ,   ω = $\sqrt{\frac{g}{l cosθ}}$

V = ω(ℓsinθ) = $\sqrt{\frac{g l}{cosθ}} sinθ$
F_av × Δt = ΔP = P₂ – P₁
mV = msinθ $\sqrt{\frac{g l}{cosθ}}$

F_av = $\frac{P_{2} - P_{1}}{Δt}$ = $\frac{2 msinθ \sqrt{\frac{g l}{cosθ}}}{π / ω}$
F_av = $\frac{2 msinθ}{π} \sqrt{\frac{g l}{cosθ}}$ × $\sqrt{\frac{g}{l cosθ}}$ ⟹ F_av = $\frac{2 mgtanθ}{π}$
L will be constant about line OC as shown in diagram