Problem 16 - Arihant AIEEE Physics

Question

A uniform brass disc of radius \(a\) and mass \(m\) is set into spinning with angular speed \(\omega_{0}\) about an axis passing through centre of disc and perpendicular to the plane of disc. If its temperature increases from \(\theta_{1}{\underline{\phantom{xx}}}^{\circ} \mathrm{C}\) to \(\theta_{2}^{\circ} \mathrm{C}\) without disturbing the disc, what will be its new angular speed ? (The coefficient of linear expansion of brass is \(\alpha\) ). (a) \(\omega_{0}\left[1+2 \alpha\left(\theta_{2}-\theta_{1}\right)\right]\) (b) \(\omega_{0}\left[1+\alpha\left(\theta_{2}-\theta_{1}\right)\right]\) (c) \(\frac{\omega_{0}}{\left[1+2 \alpha\left(\theta_{2}-\theta_{1}\right)\right]}\) (d) None of these

Step-by-Step Solution

Answer

The new angular speed is \(\frac{\omega_0}{1 + 2\alpha(\theta_2 - \theta_1)}\).

Step 1: Understanding the Expansion Effect

When the temperature of a disc increases, it undergoes thermal expansion, causing its radius and therefore its moment of inertia to change. The linear expansion coefficient \(\alpha\) relates the change in temperature \(\theta_{2} - \theta_{1}\) with the change in radius.

Step 2: Express the New Radius

Since the original radius is \(a\) and the temperature changes by \(\theta_2 - \theta_1\), the new radius \(a'\) is \(a' = a(1 + \alpha(\theta_2 - \theta_1))\).

Step 3: Calculate Change in Moment of Inertia

Moment of inertia \(I\) for a disc about its center is \(\frac{1}{2}ma^2\). The new moment of inertia \(I'\) when radius is \(a'\) will be \(I' = \frac{1}{2}m(a')^2 = \frac{1}{2}m(a(1 + \alpha(\theta_2 - \theta_1)))^2\).

Step 4: Simplify the New Moment of Inertia

Simplifying \(I'\), we get \[ I' = \frac{1}{2}m(a^2(1 + 2\alpha(\theta_2 - \theta_1) + \alpha^2(\theta_2 - \theta_1)^2)) \] and since \(\alpha^2\) is very small, it can be neglected, leading to \[ I' \approx \frac{1}{2}m(a^2(1 + 2\alpha(\theta_2 - \theta_1))) \].

Step 5: Apply Conservation of Angular Momentum

Angular momentum \(L\) is conserved, so initial \(L = I\omega_0 = \frac{1}{2}ma^2\omega_0\). After the expansion, \(L' = I'\omega'\) and must equal \(L\), so \(\frac{1}{2}ma^2\omega_0 = \frac{1}{2}ma^2(1 + 2\alpha(\theta_2 - \theta_1))\omega'\).

Step 6: Solve for New Angular Speed

From the equation \(\frac{1}{2}ma^2\omega_0 = \frac{1}{2}ma^2(1 + 2\alpha(\theta_2 - \theta_1))\omega'\), cancel out common terms to find the new angular speed \(\omega' = \frac{\omega_0}{1 + 2\alpha(\theta_2 - \theta_1)}\).

Key Concepts

Angular Momentum Conservation

When dealing with rotating objects, angular momentum conservation is a critical concept. Angular momentum (\( L \) ) is given by the product of the moment of inertia (\( I \) ) and the angular velocity (\( \omega \) ). In a closed system, where no external torques are acting, the angular momentum remains constant. This means that any change in the moment of inertia will lead to an inverse change in angular velocity to keep the angular momentum unchanged.
In the given exercise, the brass disc's temperature increases, causing it to expand. This changes the moment of inertia. But since no external torques act upon it, the angular momentum must be conserved. Initially, the angular momentum is \( L = I \omega_0 \) , and after expansion, it becomes \( L' = I' \omega' \) . By setting these equal, \( I \omega_0 = I' \omega' \) , we solve for the new angular velocity (\( \omega' \) ), which shows how the system adapts to preserve angular momentum as the physical properties change.

Moment of Inertia

Moment of inertia (\( I \) ) determines how much torque is required to bring about a certain angular acceleration in a rotating body. It's a rotational equivalent of mass in linear motion, affecting the dynamics of rotations.
For a uniform disc rotating about its center, the moment of inertia is given by \( I = \frac{1}{2} ma^2 \) , where \( m \) is the mass of the disc and \( a \) is its radius.
In the case of the brass disc, when the temperature increases, thermal expansion affects the radius, altering the moment of inertia. The new moment of inertia becomes \( I' = \frac{1}{2} m (a')^2 \) , where \( a' = a(1 + \alpha(\theta_2 - \theta_1)) \) . After expansion, since radius \( a' \) is larger due to the temperature change, \( I' \) correspondingly grows, affecting the disc's rotational speed to conserve angular momentum.

Coefficient of Linear Expansion

The coefficient of linear expansion (\( \alpha \) ) quantifies how much a material expands per degree change in temperature. It's crucial for understanding how temperature variations affect the sizes of objects and structures.
For a brass disc with an initial radius \( a \) , the expansion due to a temperature change from \( \theta_1 \) to \( \theta_2 \) can be calculated as \( a' = a(1 + \alpha(\theta_2 - \theta_1)) \) . Here,

• \( a' \) represents the new expanded radius.
• \( \alpha \) is a material-specific constant indicating its tendency to expand.
• The term \( \alpha(\theta_2 - \theta_1) \) represents the fractional increase in radius.

Understanding this concept is pivotal for problems like the one in the exercise because the amount by which the disc expands or contracts directly influences the change in its moment of inertia and thereby impacts other physical properties, like rotational speed, if not controlled externally.