A particle describes a horizontal circle on smooth inner surface of a cone as shown in figure. If the height of the circle above the vertex is 10 cm. The speed of the particle is (g, acceleration due to gravity = $10 m/s^2$)
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For a cone, the orbital speed $v$ depends only on the height $h$ from the vertex, not the mass.
Step 1: Understanding the Question:
For a particle moving in a horizontal circle inside a smooth cone, the normal reaction $N$ from the surface provides both the centripetal force and counteracts gravity. Step 2: Key Formula or Approach:
Let $\theta$ be the semi-vertical angle of the cone.
Vertical equilibrium: $N \sin \theta = mg$
Horizontal (centripetal) force: $N \cos \theta = \frac{mv^2}{r}$
Dividing the equations: $\tan \theta = \frac{rg}{v^2}$
Also, from geometry, $\tan \theta = \frac{r}{h}$, where $h$ is the height from vertex. Step 3: Detailed Explanation:
Equating the two expressions for $\tan \theta$:
\[ \frac{r}{h} = \frac{rg}{v^2} \]
\[ v^2 = gh \implies v = \sqrt{gh} \]
Given:
$h = 10 cm = 0.1 m$
$g = 10 m/s^2$
Substitute the values:
\[ v = \sqrt{10 \times 0.1} = \sqrt{1} = 1 m/s \] Step 4: Final Answer:
The speed of the particle is 1 m/s.
