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

A Box on the Edge Imagine a rectangular solid box resting precariously on the edge of a high platform. For a fleeting moment—just $\tau =0.01\text{ s}$—before it completely slips off, it pivots around the edge of the platform. This tiny window of time is where all the rotational magic happens. Gravity pulls down on the center of mass of the box, creating a turning effect, or torque, about the edge.

Angular Impulse To understand how fast the box starts spinning, we need to look at the concept of Angular Impulse. Just as a linear impulse (Force $\times$ Time) changes an object's linear momentum, an angular impulse (Torque $\times$ Time) changes its angular momentum.

The equation is beautifully simple:

Since the box starts from rest, the change in angular momentum is simply its final angular momentum, $I\omega$.

The torque is generated by the weight of the box, $mg$, acting at a perpendicular distance from the pivot. Since the center of mass is exactly in the middle of the box, this distance is $l/2$.

Next, we need the moment of inertia. Because the box is pivoting around its end (the edge of the platform), we use the formula for a rod rotating about its end:

Equating the angular impulse to the angular momentum gives us:

Notice how the mass $m$ cancels out entirely! Rearranging to solve for the angular velocity $\omega$:

Plugging in our known values ($g=10{\text{ m/s}}^2$, $\tau =0.01\text{ s}$, $l=0.3\text{ m}$):

Airborne Rotation Once the box slips off the edge, it enters free fall. With no more pivot to provide a torque, the angular velocity remains perfectly constant at $0.5\text{ rad/s}$. But how long does it fall?

We turn to the classic kinematics equation for an object dropped from rest:

Solving for the time of flight $t$:

To find the total angle rotated, we simply multiply the angular velocity by the time:

And there we have it! The box rotates by exactly $0.5\text{ radians}$ before hitting the ground. A perfect blend of rotational dynamics and linear kinematics.