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 equation is:

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

Substituting the given values:

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

Simplifying to find \( I \):

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

Solving for \( R^2 \):

\( R^2 = 400 \)

Therefore, \( R \) is:

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

And the distance \( D \) is:

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