Angular momentum

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Question

A solid Copper sphere is spinning about a diameter with angular velocity 10 rad/s. Due to rise in temperature, the radius of sphere increase by 1%. What will be the change in its angular velocity?

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Solution

Since no external torque is acting on the sphere, its angular momentum remains conserved.
$\begin{array}{l} ∴ Iω = L = constant \Rightarrow \frac{2}{5} {mR}^{2} . ω = constant \Rightarrow ω = c . R^{- 2} \\ ∴ \frac{Δω}{ω} = - 2 (\frac{ΔR}{R}) \Rightarrow \frac{Δω}{ω} \times 100 % = - 2 \times 1 % = - 2 % \end{array}$
So angular velocity decreases by 2%.

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Angular momentum

Remember concepts with our Masterclasses.

A solid Copper sphere is spinning about a diameter with angular velocity 10 rad/s. Due to rise in temperature, the radius of sphere increase by 1%. What will be the change in its angular velocity?

Since no external torque is acting on the sphere, its angular momentum remains conserved.∴Iω=L=constant⇒25mR2.ω=constant⇒ω=c.R−2∴Δωω=−2ΔRR⇒Δωω×100%=−2×1%=−2%So angular velocity decreases by 2%.

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