Question

A uniform solid brass sphere is rotating with angular speed $\omega_{0}$ about a diameter. If its temperature is now increased by 100⁰C. What will be its new angular speed.

(Given $\alpha_{B} = 2.0 \times 10^{- 5}\text{per}{^\circ}C$)

• A. $1.1\omega_{0}$
• B. $1.01\omega_{0}$
• C. $0.996\omega_{0}$
• D. $0.824\omega_{0}$

Answer 1

Answer: (C)

$0.996\omega_{0}$

Answer 2

Due to increase in temperature, radius of the sphere changes.

Let R₀ and R₁₀₀ are radius of sphere at 0⁰C and 100⁰C $R_{100} = R_{0}\lbrack 1 + \alpha \times 100\rbrack$

Squaring both the sides and neglecting higher terms $R_{100}^{2} = R_{0}^{2}\lbrack 1 + 2\alpha \times 100\rbrack$

By the law of conservation of angular momentum $I_{1}\omega_{1} = I_{2}\omega_{2}$

⇒ $\frac{2}{5}MR_{0}^{2}\omega_{1} = \frac{2}{5}MR_{100}^{2}\omega_{2}$

⇒$R_{0}^{2}\omega_{1} = R_{0}^{2}\lbrack 1 + 2 \times 2 \times 10^{- 5} \times 100\rbrack\omega_{2}$

⇒ $\omega_{2} = \frac{\omega_{1}}{\lbrack 1 + 4 \times 10^{- 3}\rbrack} = \frac{\omega_{0}}{1.004} = 0.996\omega_{0}$