A particle describe the horizontal circus of radius r on the smooth surface of an inverted cone as shown the height of plane of a circle above vertex is h the speed of particle should be (1) √rg (2) √gh (3) √2rg (4) √2gh?

Question

Answer

Explanation:

Circular Motion on Inverted Cone

When a particle moves on the smooth surface of an inverted cone of radius R and height H, the motion can be considered as a combination of two motions, i.e., horizontal and vertical.

The horizontal motion is uniform circular motion with radius R, and the vertical motion is uniform acceleration with acceleration g.

Derivation of Speed of Particle

Let v be the speed of the particle moving on the horizontal surface of the inverted cone.

The force acting on the particle is the weight of the particle, which is mg, where m is the mass of the particle and g is the acceleration due to gravity.

The horizontal component of the weight of the particle provides the centripetal force required for circular motion. Hence,

mv²/R = mg sin θ

where θ is the angle between the vertical and the radius of the circle.

sin θ = h / √(R² + h²)

Substituting sin θ in the above equation, we get

v² = gR√(R² + h²) / (R² - h²)

v = √gR√(R² + h²) / (R² - h²)

On simplifying, we get

v = √(gR² / (R - h)(R + h))

v = √(gR² / (R² - h²))

v = √(g / ((R² / h²) - 1))

Since H = R² / H, substituting this value, we get

v = √(gH / (H - h))

Therefore, the speed of the particle should be √(gH / (H - h)).

Answer:

The speed of the particle should be (2) √gh.

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