An iron sphere rotates about a diameter. With α = 1 0 − 5 ∘ C − 1 and temperature rise 10 0 ∘ C , its moment of inertia increases by:

I ∝ R 2 at fixed mass, so I I ′ = ( 1 + α Δ T ) 2 = ( 1.001 ) 2 = 1.002001 . Increase is 0.2001% , rounded to B = 0.2\%.

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PhysicsHeat & ThermoThermal ExpansionMedium2 minQB

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An iron sphere rotates about a diameter. With $α = 1 0^{- 5}^{\circ} C^{- 1}$ and temperature rise $10 0^{\circ} C$ , its moment of inertia increases by:

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Answer:

Solution:

$I \propto R^{2}$ at fixed mass, so $\frac{I ^{'}}{I} = (1 + α Δ T)^{2} = (1.001)^{2} = 1.002001$ . Increase is $0.2001%$ , rounded to $B =$ 0.2\%.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

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