A uniform metal solid sphere is rotating with angular speed $\omega_0$ about diameter. If the temperature is…

A uniform metal solid sphere is rotating with angular speed $\omega_0$ about diameter. If the temperature is raised by $50^{\circ} \mathrm{C}$, the angular speed will be [given a metal $=20 \times$ $\left.10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right]$
  1. $0.95 \omega_0$
  2. $0.96 \omega_0$
  3. $0.98 \omega_0$
  4. $\omega_0$ (Angular velocity is same)

Solution

$\Delta \mathrm{T}=50^{\circ} \mathrm{C}, \alpha=20 \times 10^{-5} \mathrm{C}^{-1}$ $\therefore \quad$ By conservation of angular momentum, $\begin{aligned} & \mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \\ & \Rightarrow\left(\frac{2}{5} \mathrm{mr}_1^2\right) \omega_0=\left(\frac{2}{5} \mathrm{mr}_1^2\right)(1+\alpha \Delta \mathrm{T})^2 \omega_2 \\ & \Rightarrow \omega_0=\left(1+20 \times 10^{-5} \times 50\right)^2 \omega_2 \\ & \therefore \omega_2=0.98 \omega_0 \end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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