A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time $$\tau = 0.01$$ s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

The rotation of the box is determined by the angular impulse it receives as it slips off the platform.

1. Calculation of Angular Velocity ($$\omega$$)

When the box slips off the edge, the torque is provided by gravity acting at its center of mass ($l/2$) relative to the edge of the platform. Using the impulse-momentum theorem for rotation:

$$\text{Angular Impulse} = \Delta L$$

$$\tau \cdot \Delta t = I\omega$$

Where:

• Torque $$\tau = mg \frac{l}{2}$$
• Moment of Inertia about the edge $$I = \frac{ml^2}{3}$$
• Time interval $$\Delta t = 0.01 \text{ s}$$

Substituting the values:

$$mg \frac{l}{2} \times 0.01 = \frac{ml^2}{3} \omega$$

$$\omega = \frac{3g \times 0.01}{2l}$$

Given $$l = 0.3 \text{ m}$ and $g = 10 \text{ m/s}^2$$:

$$\omega = \frac{3 \times 10 \times 0.01}{2 \times 0.3} = \frac{0.3}{0.6} = 0.5 \text{ rad/s}$$

2. Time of Flight ($t$)

The box falls from a height $$h = 5 \text{ m}$$. The time taken to hit the ground is:

$$t = \sqrt{\frac{2h}{g}}$$

$$t = \sqrt{\frac{2 \times 5}{10}} = 1 \text{ s}$$

3. Total Angle of Rotation ($$\theta$$)

Since the box rotates at a constant angular velocity $\omega$ during its fall:

$$\theta = \omega \cdot t$$$$\theta = 0.5 \times 1$$

$$\boxed{\theta = 0.5 \text{ radian}}$$