Calculating Rope Tension On A Pulley: A Physics Problem

Hey guys, ever found yourself staring at a physics problem that seems like a tangled mess? Let's break down one of those problems today: calculating the tension in a rope wrapped around a pulley. This is a classic physics scenario, and understanding it involves a few key concepts. So, grab your thinking caps, and let's dive in!

Understanding the Problem

So, the question is: a string is wrapped around a pulley with a radius of 10 cm, and a 0.4 kg load is attached to the end, causing the pulley to rotate. Given that the moment of inertia of the pulley is 4 x 10⁻³ kg m², what is the tension in the string? The options provided are:

• A. 0.14 N
• B. 0.57 N
• C. 2.0 N
• D. 14.0 N

This problem involves rotational motion and Newton's Second Law. We need to figure out how the force of the weight pulling down on the string translates into the tension in the string, considering the pulley's resistance to rotation (its moment of inertia).

Breaking Down the Concepts

Before we jump into calculations, let's refresh some fundamental physics concepts that are crucial for solving this problem. These concepts act as the building blocks for our solution, ensuring we understand not just the 'how' but also the 'why' behind each step.

First, let's consider Tension. In physics, tension is the pulling force transmitted axially by means of a string, cable, chain, or similar one-dimensional continuous object, or by each end of a rod or truss. It's the force that's transmitted through the rope when it is pulled tight by forces acting from opposite ends. In our pulley system, the tension is the force exerted by the rope on the pulley rim, opposing the gravitational force acting on the hanging mass. The magnitude of tension is consistent throughout the rope if we assume the rope is massless and there are no other forces acting along its length.

Now, let's delve into Torque. Torque is the rotational equivalent of linear force. It’s what causes an object to rotate. Think of it as a twisting force. Torque (${\tau}$) is calculated as the product of the force applied (F) and the distance from the axis of rotation (r), often referred to as the lever arm. Mathematically, this is expressed as ${ \tau = rF }$. In our scenario, the tension in the rope exerts a torque on the pulley, causing it to rotate. The radius of the pulley acts as the lever arm, so the greater the radius, the greater the torque exerted by the same tension force.

Next, we have Moment of Inertia. Moment of inertia (I) is an object's resistance to changes in its rotational motion, analogous to mass in linear motion. It depends on both the mass of the object and how that mass is distributed relative to the axis of rotation. A higher moment of inertia means it's harder to start or stop the object's rotation. For our pulley, the moment of inertia is a given value, and it plays a crucial role in determining how the pulley responds to the torque applied by the tension in the rope.

Finally, let's discuss Newton's Second Law for Rotational Motion. Just as Newton's Second Law (F = ma) relates force, mass, and acceleration in linear motion, there’s a rotational equivalent. It states that the net torque acting on an object is equal to the product of the object's moment of inertia (I) and its angular acceleration (α), expressed as ${ \tau = I\alpha }$. This is a cornerstone principle for our problem because it directly links the torque exerted by the tension in the rope to the pulley's resulting angular acceleration. Understanding this law allows us to connect the forces and torques to the motion of the pulley system.

Identifying Key Variables

Alright, before we crunch any numbers, let's identify the key variables we're working with. This helps us organize our thoughts and see how each piece of information fits into the puzzle. It’s like gathering all the ingredients before you start cooking – ensures a smoother process!

First, we have the Radius of the Pulley (r), which is given as 10 cm. However, we need to convert this to meters because the standard unit for calculations in physics is meters. So, 10 cm becomes 0.1 meters. Keeping units consistent is crucial in physics calculations to avoid errors. This radius is particularly important because it relates the linear tension force to the torque acting on the pulley.

Then, we have the Mass of the Load (m), which is 0.4 kg. This mass is hanging from the string and experiencing the force of gravity, which is the driving force behind the pulley's rotation. The weight of this mass (force due to gravity) directly influences the tension in the string.

We are also given the Moment of Inertia of the Pulley (I), which is 4 x 10⁻³ kg m². This value tells us how resistant the pulley is to changes in its rotational speed. A higher moment of inertia means the pulley is more difficult to start or stop rotating. This is a critical parameter in our calculations because it links the torque applied to the resulting angular acceleration.

Lastly, we're looking to find the Tension in the String (T). This is our unknown variable, the one we're trying to calculate. The tension is the force exerted by the string on the pulley and is the key connection between the linear force of the hanging mass and the rotational motion of the pulley. Understanding how these variables relate to each other is the key to solving this problem efficiently.

Step-by-Step Solution

Okay, let's get to the exciting part – solving the problem! We'll break it down into manageable steps so you can follow along easily. Think of it as building a bridge; each step is a crucial component that leads us to the final answer.

First, we need to Calculate the Weight of the Load (W). The weight is the force exerted on the mass due to gravity. We calculate it using the formula W = mg, where m is the mass (0.4 kg) and g is the acceleration due to gravity (approximately 9.8 m/s²). So, W = 0.4 kg * 9.8 m/s² ≈ 3.92 N. This weight is the force pulling down on the string, and it's the primary driver of the pulley's rotation.

Next, we Apply Newton's Second Law for Rotation. We know that the net torque (τ) acting on the pulley is equal to the moment of inertia (I) times the angular acceleration (α): τ = Iα. The torque is also related to the tension in the string (T) and the radius of the pulley (r): τ = Tr. So, we can write Tr = Iα. This equation links the tension in the rope, the pulley's properties, and its rotational motion.

Now, let's Relate Linear and Angular Acceleration. The linear acceleration (a) of the load is related to the angular acceleration (α) of the pulley by the equation a = rα. This is a key connection because it allows us to bridge the gap between the linear motion of the hanging mass and the rotational motion of the pulley.

We then Apply Newton's Second Law for the Load. For the hanging mass, the net force acting on it is the difference between its weight (W) and the tension in the string (T): F_net = W - T. According to Newton's Second Law, F_net = ma, so we have W - T = ma. This equation tells us how the tension in the string affects the acceleration of the hanging mass.

Next, we Solve for Tension (T). We now have two equations: Tr = Iα and W - T = ma. We also know that a = rα. We can rearrange these equations to solve for T. From a = rα, we get α = a/r. Substituting this into the torque equation gives Tr = I(a/r), which can be rearranged to T = Ia/r². Now, substituting W - T = ma into this, we get T = W - ma. We can then substitute a = Tr²/I into W - T = ma to get W - T = m(Tr²/I). Solving for T gives us T = W / (1 + mr²/I).

Finally, we Substitute the Values and Calculate. Plugging in the values we have: W = 3.92 N, m = 0.4 kg, r = 0.1 m, and I = 4 x 10⁻³ kg m², we get T = 3.92 / (1 + 0.4 * (0.1)² / (4 x 10⁻³)) = 3.92 / (1 + 0.004 / 0.004) = 3.92 / 2 ≈ 1.96 N. So, the tension in the string is approximately 1.96 N, which is closest to option C (2.0 N).

Detailed Calculation

Okay, let's break down the calculation process a little further. Sometimes seeing the nitty-gritty details can really help solidify understanding. We've already laid out the steps, but now let's zoom in on the math itself. It’s like looking at a recipe – the instructions are great, but understanding the measurements and how they combine is key to a perfect dish!

First, we Calculate the Weight (W). We used the formula W = mg, where m = 0.4 kg and g ≈ 9.8 m/s². So, W = 0.4 kg * 9.8 m/s² = 3.92 N. This is straightforward, but crucial – the weight is the initial force that gets everything moving.

Now, let's get into the Torque Equation. We know τ = Tr and τ = Iα. So, Tr = Iα. This equation is a cornerstone, connecting tension and rotational motion. We'll come back to this.

Next, we need to Relate Linear and Angular Acceleration using a = rα. This is where we link the linear world of the falling mass to the rotational world of the pulley. Rearranging this, we get α = a/r.

Now, let's talk about Newton's Second Law for the Load: W - T = ma. This equation describes the forces acting on the hanging mass. The net force is the difference between the weight pulling down and the tension pulling up.

Here comes the Algebraic Manipulation. This is where things get a bit more involved, but stick with me! We want to solve for T. We can substitute α = a/r into Tr = Iα to get Tr = I(a/r). Rearranging, we have T = Ia/r². We also have W - T = ma. We can substitute a from the angular acceleration equation into this, but let’s take a slightly different route this time to show another way.

Let's rearrange W - T = ma to get a = (W - T) / m. Now we substitute this a into T = Ia/r²: T = I((W - T) / m) / r². This looks complex, but we're just substituting one equation into another.

Now, let's Simplify the Equation and Solve for T. Multiplying both sides by r² gives Tr² = I((W - T) / m). Multiplying both sides by m gives Tmr² = I(W - T). Expanding the right side, Tmr² = IW - IT. Now, let’s get all the T terms on one side: Tmr² + IT = IW. Factoring out T, we get T(mr² + I) = IW. Finally, we solve for T: T = IW / (mr² + I).

This is a slightly different form than we derived earlier, but it's mathematically equivalent. It’s always good to see different routes to the same destination!

Now, let's Plug in the Values: T = (4 x 10⁻³ kg m² * 3.92 N) / (0.4 kg * (0.1 m)² + 4 x 10⁻³ kg m²). Breaking it down: The numerator is (0.004 * 3.92) = 0.01568. The denominator is (0.4 * 0.01) + 0.004 = 0.004 + 0.004 = 0.008.

Finally, Calculate T: T = 0.01568 / 0.008 ≈ 1.96 N. Again, this is very close to 2.0 N, which corresponds to option C.

Common Mistakes to Avoid

Solving physics problems can be tricky, and it's easy to stumble if you're not careful. Let's highlight some common pitfalls students often encounter when tackling problems like this. Being aware of these mistakes can save you a lot of headaches and help you ace those exams!

First up is Unit Conversion Errors. This is a classic blunder. In our pulley problem, the radius was given in centimeters (cm), but we needed to convert it to meters (m) for consistent calculations. Forgetting this conversion can throw off your entire answer. Always double-check your units and make sure they're in the standard SI units (meters, kilograms, seconds) before plugging them into equations. It's a small step that makes a huge difference!

Another frequent mistake is Incorrectly Applying Newton's Laws. Newton's laws are fundamental to mechanics problems, but applying them incorrectly can lead to wrong answers. For instance, in our problem, you need to consider Newton's Second Law for both linear motion (for the hanging mass) and rotational motion (for the pulley). Forgetting one or misapplying the equations (F = ma or τ = Iα) can derail your solution. Always make sure you're applying the right law to the right object and in the correct direction.

Then, there’s the issue of Mixing Up Torque and Force. Torque is the rotational equivalent of force, but they're not the same thing. Torque causes rotation, while force causes linear acceleration. In the pulley problem, the tension in the string exerts a torque on the pulley, causing it to rotate. Confusing these concepts can lead to incorrect calculations. Remember, torque involves both force and the distance from the axis of rotation (lever arm).

Another common pitfall is Forgetting the Relationship Between Linear and Angular Quantities. In problems involving rotational motion, linear and angular quantities are closely related. For example, linear acceleration (a) and angular acceleration (α) are related by a = rα, where r is the radius. Forgetting this relationship or misapplying it can make your calculations go haywire. Always ensure you're correctly translating between linear and angular variables.

Finally, Algebraic Errors are a significant source of mistakes. Physics problems often involve multiple steps of algebraic manipulation, and it's easy to make a mistake along the way. A simple sign error or misapplication of the order of operations can throw off your entire calculation. Take your time, write out each step clearly, and double-check your work. It’s like proofreading an essay – catching those little errors can make a big difference!

Real-World Applications

You might be thinking,