A thin uniform rod of length 2 m, cross sectional area 'A' and density 'd' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity ω. If value of ω in terms of its rotational kinetic energy E is √((α E)/(A d)) then value of α is ______________.

Solution

Physics sir · 4 steps

Write the mass as volume times density, put it into the centre-axis moment of inertia, and read off ω.

1. Mass of the rod. Volume is A × L = 2A, so the mass is m = 2Ad.
2. Moment of inertia. For an axis through the centre and perpendicular to the rod, I = mL²/12 = (2Ad × 4)/12 = 2Ad/3.

   E = 1/2Iω² = 1/2 × 2Ad/3 ω² = (Ad ω²)/3

3. Solve for ω. Rearranging gives ω² = 3E/Ad, so ω = √(3E/Ad).
4. So. Comparing with √((α E)/Ad) gives α = 3.

The common slip: 0.75. Using the end-axis formula I = mL²/3 instead of mL²/12 gives I = 8Ad/3 and α = 0.75, not 3.

The idea to keep. When a rod is described by area and density, first write m = ALd. Then pick the moment of inertia for the right axis: mL²/12 about the centre, mL²/3 about an end.

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