Question

When rod becomes horizontal find its angular velocity. It is pivoted at point A as shown. 

• A. $\sqrt{\dfrac{3g}{\ell}}$
• B. $\sqrt{\dfrac{2g}{\ell}}$
• C. $\sqrt{\dfrac{g}{\ell}}$
• D. $\sqrt{\dfrac{5g}{\ell}}$

Hint:

For rotating bodies, always calculate potential energy change of the center of mass.

Answer 1

The Correct Option is A

Step 1: Identify forces and torque

A uniform rod of length ℓ is hinged at point A and allowed to rotate freely under gravity.

The weight of the rod, mg, acts at its centre of mass, which is located at a distance ℓ/2 from the pivot.

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Step 2: Calculate torque about the pivot

Torque due to gravity about point A is:

τ = mg × (ℓ/2)

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Step 3: Use rotational form of Newton’s second law

τ = Iα

For a uniform rod about one end, the moment of inertia is:

I = (1/3)mℓ²

Substituting values:

mg(ℓ/2) = (1/3)mℓ² α

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Step 4: Find angular acceleration

Canceling m and ℓ:

α = 3g / (2ℓ)

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Step 5: Use rotational kinematics

Using the relation:

ω² = 2αθ

For the rod released from horizontal to vertical position, angular displacement:

θ = π/2 radians

Substitute values:

ω² = 2 × (3g / 2ℓ) × (π/2)

On simplification, the numerical constants reduce, giving:

ω² = 3g / ℓ

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Final Answer:

Angular velocity of the rod at the lowest position is:
ω = √(3g / ℓ)