Animated Solution for Physics - Rotational Motion: A solid disc of radius $20\text{ cm}$ and mass $10\text{ kg}$ is rotating with an angular velocity of $600\text{ rpm}$, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in $10\text{ s}$ is $\_\_\_\_\pi \times 10^{-1}\text{ N-m}$.

The problem of stopping a massive, rapidly spinning object is a classic application of rotational dynamics. Imagine a heavy solid disc, much like a miniature merry-go-round, spinning freely. Our mission is to determine the exact retarding torque required to bring this disc to a complete halt within a specific timeframe.

Let's break down the physics behind this deceleration and see how Newton's laws apply to the rotational world.

Before we can calculate the force required to stop the disc, we must understand how much the disc resists any change to its state of rotation. This resistance is quantified by its moment of inertia ($I$).

For a uniform solid disc rotating about an axis perpendicular to its plane and passing through its center, the moment of inertia is given by the formula:

We are given the mass $m=10\text{ kg}$ and the radius $R=20\text{ cm}$. It is crucial to convert the radius into standard SI units (meters) to avoid calculation errors. Thus, $R=0.2\text{ m}$.

Substituting these values into our formula:

This value, $0.2{\text{ kg m}}^2$, represents the disc's rotational inertia.

Next, we need to analyze the disc's motion. The initial angular velocity (${\omega }_0$) is given as $600\text{ rpm}$ (revolutions per minute). To use this in our standard physics equations, we must convert it to radians per second.

Since one revolution equals $2\pi$ radians and one minute equals $60$ seconds, the conversion is:

We want the disc to come to a complete stop, which means the final angular velocity (${\omega }_f$) will be $0\text{ rad/s}$. The time ($t$) allowed for this deceleration is $10\text{ s}$.

Using the first equation of rotational kinematics, we can find the angular acceleration ($\alpha$):

Substituting our known values:

The negative sign perfectly aligns with our physical intuition: it indicates a retardation or deceleration, meaning the angular acceleration is directed opposite to the rotation.

Now for the grand finale. To find the retarding torque ($\tau$), we apply Newton's second law for rotation, which states that the net torque is the product of the moment of inertia and the angular acceleration:

We use the absolute value of $\alpha$ because we are interested in the magnitude of the retarding torque. Plugging in the values we derived:

To match the specific format requested by the problem ($\text{\_\_\_\_ }\pi \times 10^{-1}\text{ N-m}$), we can rewrite our result using scientific notation:

By directly comparing our calculated torque with the given expression, it is crystal clear that the missing integer value is exactly 4.

This problem beautifully illustrates how mass distribution, kinematics, and rotational dynamics intertwine to describe the physical world!