Question

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of $25\pi$ Nm for 40s, the speed increases to 2100 rpm. The diameter of the disk is ______ m.

Hint:

- For rotational motion problems, always convert rpm to rad/s for consistency. - The moment of inertia formula depends on the axis of rotation (e.g., \(\frac{1}{4}MR^2\) for a disk rotating about its diameter). - Use the kinematic equation \(\omega = \omega_0 + \alpha t\) to relate angular acceleration and time.

Answer 1

Correct Answer: 40

The given equation is:

\( \tau \, dt = I \Delta w \)

Substitute the given values:

\( 25\pi \times 40 = I(300) \times \frac{2\pi}{60} \)

Simplify to find \( I \):

\( I = \frac{25 \times 60 \times 40}{300 \times 2} = 100 = \frac{M R^2}{4} \)

Now, solving for \( R^2 \):

\( R^2 = 400 \)

Thus, the value of \( R \) is:

\( R = 20 \, \text{m} \)

And the distance \( D \) is:

\( D = 40 \, \text{m} \)

Answer 2

Step 1: Convert initial and final angular speeds from rpm to rad/s.
- Initial speed, \(\omega_0 = 1800 \, \text{rpm} = \frac{1800 \times 2\pi}{60} = 60\pi \, \text{rad/s}\).
- Final speed, \(\omega = 2100 \, \text{rpm} = \frac{2100 \times 2\pi}{60} = 70\pi \, \text{rad/s}\).

Step 2: Calculate angular acceleration (\(\alpha\)) using torque and moment of inertia.
- Torque, \(\tau = 25\pi \, \text{Nm}\).
- Moment of inertia for a thin disk rotating about its diameter: \( I = \frac{1}{4}MR^2 = \frac{1}{4} \times 1 \times R^2 = \frac{R^2}{4}\).
- Angular acceleration: \( \tau = I\alpha \quad \Rightarrow \quad 25\pi = \frac{R^2}{4} \alpha \quad \Rightarrow \quad \alpha = \frac{100\pi}{R^2}\).

Step 3: Relate angular acceleration to the change in angular velocity.
\( \omega = \omega_0 + \alpha t \quad \Rightarrow \quad 70\pi = 60\pi + \left(\frac{100\pi}{R^2}\right) \times 40\).
Simplify: \( 10\pi = \frac{4000\pi}{R^2} \quad \Rightarrow \quad R^2 = 400 \quad \Rightarrow \quad R = 20 \, \text{m}\).
- Diameter, \(D = 2R = 40 \, \text{m}\).