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Grade 11Mechanics

A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume ) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in liquid of density σ (> ρ).

Radhika Batra

12 Years agoGrade 11

1 Answer

Kevin Nash

12 Years ago

Hello Student,

Please find the answer to your question

Note :

For the wooden stick-mass system to be in stable equilibrium the center of gravity of stick-mass system should be lower than the center of buoyancy. Also in equilibrium the centre of gravity (N) and the centre of buoyancy (B) lie on the same vertical axis.

The above condition 1 will be satisfied if the mass is towards the lower side of the stick as shown in the figure The two forces will create a torque which will bring the stick-mass system in the vertical position of the stable equilibrium.

Let ℓ be the length of the stick immersed in the liquid.

Then OB = ℓ / 2

For vertical equilibrium

F_G = F_B ⇒ (M+ m) g = F_B

⇒ π R² L ρ g + mg = π R² ℓ σ g

ℓ = πR² Lρ + m / πR²σ …(1)

Let the distance of the center of mass N of the (rod + mass) system from the origin O be ON = y. Then

y = My₁ + my₂ / M + m

Since mass m is at O, the origin, therefore y₂ = 0

∴ y = M(L/2) + m x 0 / M + m = ML / 2 (M + m)

= (π R² L ρ) L / 2 (π R² L ρ + m) …(2)

Therefore for stable equilibrium

ℓ / 2 > y

∴ πR² Lρ + m / 2(πR² L σ) > (π R² Lρ) L / 2 (π R² Lρ + m)

⇒ m ≥ π R² L (√ρσ - ρ)

∴ Minimum value of m is π R² L (√ρσ - ρ)

Thanks
Kevin Nash
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