One end of a straight uniform 1m long bar is pivoted on horizontal table. It is released from rest when it makes an angle 30º from the horizontal (see figure). Its angular speed when it hits the table is given as √n s⁻¹, where n is an integer. The value of n is _________.

Solution

Physics sir · 4 steps

The pivot does no work, so the potential energy the centre loses all turns into rotational KE about the pivot.

1. Height lost. Gravity acts at the centre of mass, which sits at L/2 along the bar. At 30° it is L/2 sin 30° = 1/2 × 1/2 = 0.25 m above the table. That is how far it falls.
2. Rotational KE. The bar turns about its fixed end, so use the moment of inertia about that end: I = mL²/3. All of its kinetic energy is 1/2Iω².

   mg L/2 sin 30° = 1/2(mL²/3)ω²

3. Put in numbers. With L = 1 m and g = 10 m s⁻², the mass cancels. This gives 10 × 0.25 = ω²/6, so ω² = 15.
4. So. ω = √15 s⁻¹, so n = 15.

The common slip: 60. Using I = mL²/12 about the centre instead of mL²/3 about the pivot gives ω² = 60, not 15.

The idea to keep. For a body rotating about a fixed pivot, write mgΔ h_(cm) = 1/2Iₚᵢᵥₒₜω². Take the height drop of the centre of mass and the moment of inertia about the pivot, not about the centre.

Answer from the NTA final answer key. Solution by PrepWiser.