Question

Starting from rest at \(t=0\), a wheel undergoes a constant angular acceleration. When \(t=2.33 \mathrm{~s}\), the angular velocity of the wheel is \(4.96 \mathrm{rad} / \mathrm{s}\). The acceleration continues until \(t=23.0 \mathrm{~s}\), when it abruptly ceases. Through what angle does the wheel rotate in the interval \(t=0\) to \(t=46.0 \mathrm{~s}\) ?

Short answer

The wheel rotates through an angle of approximately 1687 rad from t=0 to t=46.0 s.

Step-by-step solution

Calculate Angular Acceleration

Given that the wheel starts from rest, the initial angular velocity (w0) is 0 rad/s. The final angular velocity (w) at t = 2.33 s is 4.96 rad/s. We can use the equation \(w = w0 + \alpha*t\) where \(\alpha\) is the angular acceleration to solve for \(\alpha\). So, \(\alpha = (w - w0) / t = 4.96 / 2.33 = 2.13 \, rad/s^2\).

Determine Angle of Rotation for t=0 to t=23.0 s

Now, we find the angle of rotation for the first 23.0s. We know that the angular acceleration is constant during this time. Therefore, we can use the equation for angular displacement (\(\theta\)) for uniformly accelerated motion: \(\theta = w0*t + 0.5*\alpha*t^2\). Substituting the known values, we obtain \(\theta = 0.5*2.13*23.0^2 = 558.045 \, rad\).

Determine Angle of Rotation for t=23.0 s to t=46.0 s

For the period from 23.0 s to 46.0 s, we know that the wheel is not accelerating, so it's rotating at a uniform rate (the final angular velocity from the accelerated period). Hence, we can use the equation \(\theta = w*t\) to find the angle of rotation during this period, where w is the angular velocity at 23.0 s. The angular velocity is given by the final angular velocity from the accelerated period and \(w = w0 + \alpha*t = 0 + 2.13*23.0 = 48.99 \, rad/s\). So, \(\theta = 48.99*(46.0-23.0) = 1128.77 \, rad\).

Calculate Total Angle of Rotation from t=0 to t=46.0 s

Finally, add the angles of rotation calculated in steps 2 & 3 to find the total angle of rotation from t=0 to t=46.0 s. Hence, the total angle of rotation \(\theta_total = 558.045 + 1128.77 = 1686.815 \, rad\).

Key concepts

Angular Acceleration

Angular acceleration is all about how quickly something speeds up as it spins. In our exercise, we start with a wheel that isn't moving at all. This means its initial angular velocity is just 0 rad/s.

But as time goes by, the wheel begins to pick up speed due to a constant force applied. This change in speed is where angular acceleration comes into play. We calculate angular acceleration using the formula:

• \( \alpha = \frac{w - w0}{t} \)

Where:

• \(w\) is the final angular velocity,
• \(w0\) is the initial angular velocity, and
• \(t\) is the time it takes to reach that speed.

The angular acceleration in our problem, for the first 2.33 seconds, is

• \(\alpha = \frac{4.96}{2.33} = 2.13 \, \text{rad/s}^2\)

This means that every second, the wheel's velocity increases by 2.13 rad/s.

Understanding this concept is crucial when dealing with situations where anything rotates and gradually gains speed. Knowing the rate at which speed changes can help predict future movement and behavior, whether it's a bike wheel or a spinning planet.

Angular Velocity

Angular velocity tells us how fast something is spinning around. Think of it like the speedometer in a car, but for rotation.

In the exercise, after the initial 2.33 seconds, our wheel reaches an angular velocity of 4.96 rad/s. This number gives us an idea of the rotational speed the wheel achieves due to the constant acceleration. When dealing with angular velocity, we use the same units as we do for angular displacement, "radians," along with time. Radians measure the angle and essentially provide a sense of distance covered around a circle. After the wheel stops accelerating at 23.0 seconds, it continues to rotate at whatever speed it reached at that point. The angular velocity helps us determine how much further the wheel will spin until the 46.0-second mark. Angular velocity can be constant or it can change, depending on forces applied. For the last part of our problem, the angular velocity is constant because the acceleration stops. This means for 23 seconds, the wheel spins at the same speed. Understanding this concept is important, as it describes the nature of continuous motion and how forces applied can change these speeds over time.

Angular Displacement

Angular displacement measures how much an object has rotated in radians, from start to endpoint. It's akin to measuring how far someone has walked but in a circular path.

In our exercise, we calculate angular displacement by using the formula:

• \( \theta = w0 \, t + 0.5 \, \alpha \, t^2 \)

during the time when the wheel is accelerating, and

• \( \theta = w \, t \)

when it is rotating uniformly. First, from 0 to 23 seconds, with constant acceleration, we compute:

• \(\theta = 0.5 \cdot 2.13 \cdot (23.0)^2 = 558.045 \, \text{rad} \)

This tells us how much the wheel has turned when it's speeding up.

Then, from 23 to 46 seconds, with a constant angular velocity of 48.99 rad/s, we find:

• \( \theta = 48.99 \cdot (46.0-23.0) = 1128.77 \, \text{rad} \)

The total angular displacement is the sum of both periods:

• \(\theta_{total} = 558.045 + 1128.77 = 1686.815 \, \text{rad} \)

This total tells us how much the wheel has rotated from the initial to the final time considered in the problem. Angular displacement is essential in understanding how much movement has occurred in rotational systems.