Animated Solution for Physics - Rotational Motion: Moment of inertia of a body about a given axis is $1.5{\text{ kg m}}^2$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1200\text{ J}$, the angular acceleration of $20{\text{ rad/s}}^2$ must be applied about the axis for a duration of

Imagine you are standing next to a massive, heavy flywheel. It's perfectly balanced, mounted on frictionless bearings, and currently, it is completely motionless. This flywheel has a specific resistance to being spun up, a property we call the moment of inertia, denoted by $I$. In our scenario, this moment of inertia is exactly $1.5{\text{ kg m}}^2$.

Our objective is to inject a massive amount of energy into this system. Specifically, we want the flywheel to possess a rotational kinetic energy of $1200\text{ J}$. To achieve this, we are going to apply a twisting force—a torque—that will cause the flywheel to accelerate. We are told that this applied torque results in a constant angular acceleration, $\alpha$, of $20{\text{ rad/s}}^2$.

The core question we must answer is: How long must we sustain this acceleration to reach our energy target?

To understand how to reach our goal, we first need to understand what the goal looks like mathematically. The energy stored in a rotating object is known as rotational kinetic energy. The formula for this is one of the most beautiful and fundamental equations in rotational mechanics:

This equation tells us that the kinetic energy depends on two things: how hard the object is to spin (its moment of inertia, $I$) and how fast it is spinning (its angular velocity, $\omega$).

We already know the target kinetic energy ($1200\text{ J}$) and the moment of inertia ($1.5{\text{ kg m}}^2$). If we wanted to, we could use this equation right now to calculate the exact angular velocity the flywheel needs to reach. However, we can be more efficient by combining our equations before we start crunching the numbers.

We know where we are starting (at rest, so initial angular velocity ${\omega }_0=0$) and we know how fast we are accelerating ($\alpha =20{\text{ rad/s}}^2$). We need a way to connect this acceleration over time to the final angular velocity.

Because the angular acceleration is constant, we can rely on the rotational equivalent of the kinematic equations we use for linear motion. The first equation of rotational kinematics is perfect for this job:

This equation simply states that the final angular velocity is equal to the initial angular velocity plus the change in velocity (which is acceleration multiplied by time). Since our flywheel starts from a dead stop, ${\omega }_0$ is zero. This simplifies our kinematic journey to a very clean relationship:

This tells us exactly how fast the flywheel is spinning at any given second $t$.

Now comes the elegant part of physics: synthesizing different concepts into a single, powerful tool. We have an equation for energy that requires $\omega$, and we have an equation that tells us what $\omega$ is at any time $t$. Let's substitute the kinematic equation into the energy equation:

This is our master equation. It directly links the time we apply the acceleration to the total energy stored in the flywheel. It bypasses the need to calculate the intermediate angular velocity entirely.

With our master equation ready, it is time to substitute the raw values provided in the problem. We know $KE=1200$, $I=1.5$, and $\alpha =20$. Let's plug them in carefully:

The first step in simplifying this is to handle the squared term. Remember that both the $20$ and the $t$ must be squared:

Now, let's multiply the constants on the right side of the equation. We can take half of $1.5$, which is $0.75$, and multiply that by $400$. Alternatively, we can take half of $400$, which is $200$, and multiply that by $1.5$. Both paths lead to the same result:

The physics problem has now completely transformed into a basic algebra problem. We want to isolate $t^2$, so we divide both sides of the equation by $300$:

Finally, to find the time $t$, we take the square root of both sides. Since time must be a positive value in this context, we only consider the positive root:

The math tells us that it takes exactly two seconds to reach our energy goal. But let's take a moment to appreciate what this means physically. In just two seconds, a constant twisting force has pumped $1200\text{ Joules}$ of energy into the system.

If you were to calculate the final angular velocity ($\omega =20\times 2=40\text{ rad/s}$), you would see that the flywheel is spinning quite rapidly. This entire process is a perfect demonstration of the work-energy theorem applied to rotational motion: the work done by the torque over those two seconds is exactly equal to the $1200\text{ J}$ of kinetic energy gained by the flywheel. Understanding this connection between time, acceleration, and energy is crucial for designing everything from car engines to advanced energy storage systems!