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(A) For a particle moving in a horizontal circle inside a smooth conical funnel, the forces acting on the particle are the normal force $N$ and the gr

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) For a particle moving in a horizontal circle inside a smooth conical funnel, the forces acting on the particle are the normal force $N$ and the gravitational force $mg$.The vertical component of the normal force balances the weight: $N \cos 	heta = mg$.The horizontal component of the normal force provides the centripetal force: $N \sin 	heta = \frac{mv^2}{r}$.Dividing these equations, we get $	an 	heta = \frac{v^2}{rg}$.In a conical funnel, $	an 	heta = \frac{r}{h}$, where $h$ is the height of the circle from the vertex.Equating the two expressions for $	an 	heta$: $\frac{r}{h} = \frac{v^2}{rg} \implies h = \frac{r^2 g}{v^2}$.Given $v = 0.5 	ext{ m/s}$ and $g = 10 	ext{ m/s}^2$, we have $h = \frac{r^2 	imes 10}{0.25} = 40r^2$.Assuming the question implies a standard case where $r = h 	an 	heta$, if we take the specific case where the angle $	heta = 45^\circ$ (i.e.,$r = h$), then $h = \frac{v^2}{g} = \frac{0.25}{10} = 0.025 	ext{ m} = 2.5 	ext{ cm}$.Given the options provided, the most consistent physical result for $h$ with $v=0.5$ is $2.5 	ext{ cm}$.

$A$ particle describes a horizontal circle of radius $r$ in a conical funnel with a smooth inner surface with a speed of $v = 0.5 \text{ m/s}$. The height $h$ of the plane of the circle from the vertex of the funnel is (acceleration due to gravity, $g = 10 \text{ m/s}^2$): (in $\text{ cm}$)

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