Velocity Vector In Polar Coordinates: A Physics Problem

by TextBrain Team 56 views

Hey guys! Let's dive into a fascinating physics problem involving an object moving in polar coordinates. This kind of problem often pops up in classical mechanics and is super important for understanding how objects move in curved paths. We'll break down the concepts, the math, and the solutions step by step, making sure it all clicks. So, buckle up and let's get started!

Problem Statement: Decoding the Motion

So, here’s the deal. We have an object with a mass of 4 kg scooting around on a flat surface. Its movement is described using polar coordinates – that's (r,θ){(r, \theta)} for those of you keeping score at home. The object’s position is defined by these equations:

r=5e−0.5tr^{r = 5e^{-0.5t} \hat{r}} θ=3t2{\theta = 3t^2}

What we need to figure out is the velocity vector v⃗(t){\vec{v}(t)}. Sounds intimidating? Don't sweat it! We'll tackle it together. This problem is a classic example of applying calculus and vector concepts in a physics context. The position vector gives us a way to track where the object is at any given time, and from that, we can derive the velocity vector, which tells us how the object’s position is changing over time. Understanding these concepts is crucial for anyone diving into dynamics or celestial mechanics. The cool thing about polar coordinates is how they naturally describe circular or rotational motion, which is why they're used so often in physics. So, let's break it down and see how we can find that velocity vector.

Understanding Polar Coordinates

Before we jump into the math, let's make sure we're all on the same page about polar coordinates. Instead of using the usual (x,y){(x, y)} Cartesian coordinates, polar coordinates use a radial distance r{r} and an angle θ{\theta} to pinpoint a location. Think of it like this: r{r} is how far you are from the origin, and θ{\theta} is the angle you need to turn from the positive x-axis. For our problem, we have r=5e−0.5t{r = 5e^{-0.5t}}, which means the distance from the origin changes with time, specifically decaying exponentially. Meanwhile, θ=3t2{\theta = 3t^2} tells us that the angle is increasing quadratically with time, so the object is spinning faster and faster as time goes on. Now, the velocity vector in polar coordinates isn't as straightforward as just taking the derivative of r{r} and θ{\theta}. We need to consider the unit vectors in polar coordinates, which are r^{\hat{r}} (pointing in the direction of increasing r{r}) and θ^{\hat{\theta}} (pointing in the direction of increasing θ{\theta}). These unit vectors themselves change direction as the object moves, which adds a bit of complexity but also makes things super interesting. So, with that in mind, let's roll up our sleeves and calculate the velocity vector.

Finding the Velocity Vector: Step-by-Step

Alright, let's get our hands dirty with the math and find that velocity vector. The key here is to remember that velocity is the time derivative of position. In polar coordinates, the position vector r⃗{\vec{r}} is given by:

r⃗=rr^{\vec{r} = r \hat{r}}

So, to find the velocity v⃗{\vec{v}}, we need to differentiate this with respect to time. But here's the tricky part: both r{r} and r^{\hat{r}} can change with time. This means we need to use the product rule for differentiation. The product rule, for those who need a refresher, says that the derivative of a product of two functions is the derivative of the first times the second, plus the first times the derivative of the second. In math terms:

ddt(uv)=dudtv+udvdt{\frac{d}{dt}(uv) = \frac{du}{dt}v + u\frac{dv}{dt}}

Applying this to our position vector, we get:

v⃗=dr⃗dt=ddt(rr^)=drdtr^+rdr^dt{\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(r \hat{r}) = \frac{dr}{dt} \hat{r} + r \frac{d\hat{r}}{dt}}

Now, we need to figure out what drdt{\frac{dr}{dt}} and dr^dt{\frac{d\hat{r}}{dt}} are. We already have r=5e−0.5t{r = 5e^{-0.5t}}, so finding drdt{\frac{dr}{dt}} is a straightforward derivative. As for dr^dt{\frac{d\hat{r}}{dt}}, this is where things get a bit polar-coordinate-specific. The rate of change of the radial unit vector r^{\hat{r}} is related to the angular velocity and the tangential unit vector θ^{\hat{\theta}}. We'll need to use the relationship dr^dt=θ˙θ^{\frac{d\hat{r}}{dt} = \dot{\theta} \hat{\theta}}, where θ˙{\dot{\theta}} is the time derivative of θ{\theta}. Once we nail these down, we'll plug everything back into our equation for v⃗{\vec{v}} and see the velocity vector emerge. So, let’s break it down further and calculate these derivatives.

Calculating drdt{\frac{dr}{dt}}

Okay, let's start with the easier part: calculating drdt{\frac{dr}{dt}}. We know that:

r=5e−0.5t{r = 5e^{-0.5t}}

To find drdt{\frac{dr}{dt}}, we simply differentiate this expression with respect to time t{t}. Remember the chain rule from calculus? It's going to be our best friend here. The chain rule tells us that if we have a composite function, like f(g(t)){f(g(t))}, its derivative is f′(g(t))⋅g′(t){f'(g(t)) \cdot g'(t)}. In our case, we have an exponential function, so the derivative will involve the exponential function itself and the derivative of the exponent. Differentiating 5e−0.5t{5e^{-0.5t}} with respect to t{t}, we get:

drdt=5⋅(−0.5)e−0.5t=−2.5e−0.5t{\frac{dr}{dt} = 5 \cdot (-0.5)e^{-0.5t} = -2.5e^{-0.5t}}

So, drdt{\frac{dr}{dt}} is −2.5e−0.5t{-2.5e^{-0.5t}}. Notice the negative sign here. This tells us that the radial distance r{r} is decreasing with time, which makes sense because of the e−0.5t{e^{-0.5t}} term. As time goes on, the object is getting closer to the origin. Now that we've got drdt{\frac{dr}{dt}} sorted, let's move on to the trickier part: finding dr^dt{\frac{d\hat{r}}{dt}}. This involves understanding how the unit vectors in polar coordinates change with time, which is a key concept in dealing with motion in polar coordinates. So, let's dive into that next and unlock the secrets of dr^dt{\frac{d\hat{r}}{dt}}.

Finding dr^dt{\frac{d\hat{r}}{dt}}

Now, let's tackle the slightly more challenging part: finding dr^dt{\frac{d\hat{r}}{dt}}. This is where the magic of polar coordinates really shines. Remember, r^{\hat{r}} is the unit vector pointing in the radial direction. As the object moves, the direction of r^{\hat{r}} changes, so we need to figure out how it changes with time. It turns out that the rate of change of r^{\hat{r}} is related to the angular velocity and the tangential unit vector θ^{\hat{\theta}}. The relationship we need is:

dr^dt=θ˙θ^{\frac{d\hat{r}}{dt} = \dot{\theta} \hat{\theta}}

Where θ˙{\dot{\theta}} (pronounced