A uniform solid brass sphere is rotating with angular speed ω₀ about a diameter. If its temperature is now increased by 100°C. What will be its new angular speed. (Given αb = 2.0 x 10⁻⁵ per°C )

The correct option is 0.996 ω₀.

To find the new angular speed of the rotating brass sphere, we can use the principles of conservation of angular momentum and thermal expansion.

Step 1: Conservation of Angular Momentum
Since there is no external torque acting on the rotating solid brass sphere, its angular momentum ($L$) is conserved during the temperature increase:
$I_0{\omega }_0=I\omega$
where:
• $I_0$ is the initial moment of inertia and ${\omega }_0$ is the initial angular speed.
• $I$ is the final moment of inertia and $\omega$ is the final angular speed.

Step 2: Moment of Inertia of a Solid Sphere
The moment of inertia of a uniform solid sphere of mass $M$ and radius $R$ about its diameter is given by:
$I=\frac{2}{5}M{R}^2$
Since the mass of the sphere remains constant, the relation between the initial and final states is:
$\left(\frac{2}{5}M{R}_0^2\right){\omega }_0=\left(\frac{2}{5}M{R}^2\right)\omega$
Simplifying this, we get:
$R_0^2{\omega }_0=R^2\omega$
Or, solving for the new angular speed $\omega$:
$\omega ={\omega }_0{(\frac{R_0}{R}}^2$

Step 3: Thermal Expansion of Radius
When the temperature of the sphere increases by $\mathrm{\Delta T}$, its radius expands according to the linear expansion formula:
$R=R_0(1+\alpha \mathrm{\Delta T})$
where:
• $\alpha$ is the coefficient of linear expansion ($2.0\times {10}^{-5}{\text{/ °C}}^{}$).
• $\mathrm{\Delta T}$ is the change in temperature ($100\text{°C}$).

Step 4: Calculation of the New Angular Speed
Substitute the expanded radius into the conservation of angular momentum equation:
$\omega ={\omega }_0{\left(\frac{R_0}{R_0(1+\alpha \mathrm{\Delta T})}\right)}^2$
$\omega =\frac{{\omega }_0}{{(1+\alpha \mathrm{\Delta T})}^2}$
Using the binomial approximation ${(1+x)}^{-2}\approx 1-2x$ (since $\alpha \mathrm{\Delta T}$ is much smaller than 1):
$\omega \approx {\omega }_0(1-2\alpha \mathrm{\Delta T})$

Now, calculate the value of the term $2\alpha \mathrm{\Delta T}$:
$2\alpha \mathrm{\Delta T}=2\times (2.0\times {10}^{-5})\times 100$
$2\alpha \mathrm{\Delta T}=4.0\times {10}^{-3}=0.004$

Substitute this value back into the approximation:
$\omega \approx {\omega }_0(1-0.004)$
$\omega \approx 0.996{\omega }_0$