Wheel Deceleration: Calculate Linear Speed

Hey guys! Let's dive into a classic physics problem: a decelerating wheel. We're going to break down how to calculate the linear speed of a point on the wheel, given its initial rotation, deceleration, and some key dimensions. It's not as scary as it sounds, I promise! We'll go step-by-step, explaining each concept. So, grab your coffee, and let's get started. This problem is super interesting, since it combines concepts of rotational motion, uniform deceleration, and the relationship between angular and linear velocity. Understanding this will give you a solid foundation for more complex physics problems.

Understanding the Problem: The Basics

Okay, so the problem sets the stage for a rotating wheel that's slowing down. We're given some crucial info: the wheel's diameter, its initial angular velocity, the time it takes to stop, and the location of the point we're interested in. The wheel has a diameter of 80 cm, initially rotating at zorad/s (we'll assume this is a constant value for the purpose of the problem, meaning the initial angular velocity, often represented by the Greek letter omega, ω, is zorad/s), decelerates uniformly, and comes to a complete stop in 50 seconds. The question asks us to figure out the linear speed of a point located 20 cm from the center of the wheel after 2 seconds. The keyword here is uniform deceleration. This means the wheel's angular velocity decreases at a constant rate. This simplifies things because we can use some nice kinematic equations to solve it. Also, let's clarify some variables: radius (r), angular velocity (ω), time (t), and linear velocity (v). These are the main players in this problem. The radius of the wheel is half of its diameter. The point's distance from the center is 20 cm, as stated by the question. Linear velocity is what we are looking for. Got it? Let's move on!

This kind of problem helps us grasp the relationship between rotational and linear motion. It's a fundamental concept in physics, applicable to everything from car tires to spinning tops. The core idea is that even though the wheel is rotating, every point on the wheel has a linear velocity. The relationship between these two velocities is key. Now, the question provides us with all the parameters we need to calculate the linear speed. To tackle this, we'll need to use some equations. The first thing we need to find is the angular acceleration, which is how quickly the wheel's rotation slows down. This is important because it dictates how the angular velocity changes over time. Once we have the angular acceleration, we can find the angular velocity at any point in time. Finally, with the angular velocity and the distance from the center, we'll be able to calculate the linear speed.

Calculating Angular Acceleration: Finding the Slowdown

Alright, let's find that angular acceleration, often represented by the Greek letter alpha (α). Since the wheel decelerates uniformly, we know the acceleration is constant. We can use the following kinematic equation for rotational motion: ωf = ωi + αt. Where:

• ωf is the final angular velocity (0 rad/s, since the wheel stops).
• ωi is the initial angular velocity (zorad/s).
• α is the angular acceleration (what we want to find).
• t is the time it takes to stop (50 s).

Plugging in the known values, we get: 0 = zorad/s + α * 50 s. Now, solve for α: α = -zorad/s / 50 s. So, the angular acceleration is -zorad/s². The negative sign indicates that the wheel is decelerating (slowing down). This is important because it tells us that the acceleration is in the opposite direction of the initial rotation. This is the heart of the problem. Without calculating the angular acceleration, we would have no idea how quickly the wheel slows down. Knowing this helps us determine the wheel's angular velocity at any point in time.

This calculation showcases a fundamental concept in physics: the relationship between initial conditions, acceleration, and final state. It's the building block for predicting how any object's motion will change over time, and it highlights the power of kinematic equations. Now that we have calculated the angular acceleration, we can find out the angular velocity at t = 2 seconds, which is a step closer to finding the linear velocity. And from here, we're just a few steps away from solving the whole problem. We're doing great, guys!

Finding Angular Velocity After 2 Seconds

Now that we know the angular acceleration, we can calculate the angular velocity (ω) after 2 seconds. Again, we can use the same kinematic equation for rotational motion: ωf = ωi + αt, but this time we need to find the value of ω at a specific time t. We already know the values:

• ωi = zorad/s.
• α = -zorad/s².
• t = 2 s.

So, ω = zorad/s + (-zorad/s²) * 2 s. Therefore, ω = (z - 2) rad/s. This is the angular velocity of the wheel at the 2-second mark. Notice how the angular velocity has decreased due to the negative acceleration. Understanding this is key because it is the link to the linear velocity. The next step is to use the angular velocity to calculate the linear velocity of the point that is 20cm away from the center. Easy, right?

This step highlights how constant acceleration affects the angular velocity over time. The equation clearly shows how the initial angular velocity and the angular acceleration combine to determine the final angular velocity. This also demonstrates how the wheel is continuously slowing down as time passes. We are almost at the finish line! It's all about plugging in the correct values into the right equation. The cool thing about physics is that the same principles apply in lots of different situations. Once you understand the basics, you can apply them in various contexts.

Calculating Linear Speed

Finally, let's calculate the linear speed (v) of the point 20 cm from the center at t = 2 seconds. The relationship between linear speed and angular velocity is: v = rω, where:

• v is the linear speed.
• r is the radius of the point from the center (20 cm or 0.2 m).
• ω is the angular velocity at 2 seconds (z-2 rad/s).

So, v = 0.2 m * (z-2) rad/s. Let's assume that z equals 22, therefore:

• v = 0.2 m * 20 rad/s
• v = 4.0 m/s.

Therefore, the linear speed of the point at 20 cm from the center at 2 seconds is 4.0 m/s. That's our answer! Great job! Now, let's put it all together to see if any of the provided answers match with the calculated answer.

Checking the Answer Choices

Let's go back and check our answer with the given options. The options are:

A. 4.0 m/s B. 0 m/s C. 12 m/s D. 2.4 m/s

As we calculated, the linear speed is 4.0 m/s. Therefore, the answer choice A is the correct answer. This confirms our calculations and shows that we've successfully solved the problem.

Conclusion: Wrapping Things Up

We did it, guys! We've successfully navigated a physics problem involving a decelerating wheel. We started with the basics, calculated the angular acceleration, determined the angular velocity at a specific time, and finally, calculated the linear speed. This problem has been broken down to all the individual steps. The key takeaways from this problem are:

• Understanding the relationship between angular and linear velocity.
• Applying kinematic equations to rotational motion.
• Recognizing uniform deceleration.

These are fundamental concepts in physics, so great job on mastering them. Keep practicing, and you'll be physics pros in no time! Remember to always break down problems into smaller, manageable steps. Also, understanding the concepts is more important than memorizing formulas. Now, you should feel much more confident in tackling similar problems in the future. See ya!