Physics

I’m writing this article because I felt that it was significant to understanding a good deal of physics and astronomy aspects that rely on this type of mathematics. By the very title, I’ve implied the need to apply the big bad demon known as Calculus. Had I better understood this part of calculus sooner it would have made a significant difference in my ability to handle some of the more complex physics problems. As a result, I thought I would share in hopes that this might help someone in the future.

Although this article will focus on the shortcut methods to working out some of the problems presented and the need for in-depth understanding of calculus is not needed. To make sense of everything the reader should have a handle on basic geometry and trigonometry as well as some familiarity with the following calculus topics:

Functions
Continuity and Limits
Differentials
Partial Differential Equations

So, let’s begin with the classic melting snowball problem:

A spherical snowball is melting in such a way that its radius is decreasing at rate of 0.5 cm/min. At what rate is the volume of the snowball decreasing when the radius is 3 cm. (Note the answer is a positive number).

Step 1: Identify what we have and what we’re looking for.

Radius: $3 cm$
Decreasing Rate: $0.5 cm/min$ decreasing radius
Additional Information: We also have a sphere and we know that the sphere is decreasing which means that it’s volume is also decreasing. Thus, we need the formula for the volume of a sphere since that’s what’s changing: $\frac{4}{3}\pi r^3$

Step 2: Now for the trick. One very handy piece of calculus that allows us to handle related rates problems is differentiation, specifically the differential. It is important to note that this is not exactly the same thing as taking the derivative, but that action is one process in this problem. The resulting formula for this process is:

$dy = f'(x) \cdot dx$

Step 3: To break this formula down a bit, dy (since we’re working with volume this will be dV for us) is the part you’re looking for in this problem; solving for it will give you the rate at which the volume is changing in relation to the radius of the sphere. The next part, f'(x) is the derivative of the formula that you’re working with, in this case, the formula for the area of a sphere. Finally, dx is the current value of the rate of change (or slope of the tangent line to the function at the given point, if that makes more sense) of whatever it is we happen to be looking for. In this case, it would be the rate that the radius of the sphere is changing.

Step 4: Now that we’ve established our variables and the equation needed to solve this problem we can set up the problem further.

We must find the derivative of the equation of the volume of a sphere:

$V = \frac{4}{3}\pi r^3$

$\frac{dV}{dr} = 4\pi r^2$

Now, substitute in the values into our differential:

$dV = (4\pi*(3 cm)^2) (.5 cm/min)$

Now solve:

$dV = 36\pi cm^2 \cdot 0.5 cm/min$

$dV = 18\pi cm^3/min$ or $56.55 cm^3/min$

And there we have it; when the radius of the snowball is 3 centimeters, the volume of the snowball is decreasing by $18\pi cm^3$ or $56.55 cm^3$ every minute! It must be one hot day out!

Stay tuned for a few follow up problems related to this article.

A question was posed in Physics as follows:

A 60 kg ball is tied to the end of a 50 cm long mass-less string and swung in a vertical circle. The center of the circle is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack. If the string is released at the instant the ball is at the top of the loop, how far to the right (in meters) does the ball hit the ground?

Okay, now we need to look up the key items in this question:

mass = 60 kg
radius = .5 m (50 cm)
height = 1.5 m (150 cm)
gravitational acceleration = 9.8 m/s^2
vertical circle (tossed overhead with release at the peak)
minimum speed for no slack ( t must be greater than or equal to 0)

Now, what this tells us is that we need to a) find the tangential velocity at the top of the circle, b) find the minimum tension on the rope that it won’t go slack at the top, c) the time that it takes the ball to fall back to the ground from it’s highest point (this will allow us to calculate distance on the x-axis). It is important to note that for this equation mass is extra information.

Let’s solve for the time from the top of the circle to the ground since that’s the easiest to get right away (a quick note, the peak is the height to the middle plus the radius of the circle, so, 2 meters):

$T = \sqrt{\frac{2d}{a}} \equiv T = \sqrt{\frac{2(2 m)}{9.8 m/s^2}} \equiv T = 0.64 s$

Conceptually we need to realize that the velocity of the object at the top of the circle will be related to the gravity, radius and tension. Since we want the minimum speed possible, that means the tension must be 0 or greater but not less than 0. That means that the tension goes out of the problem because we’re just dealing with gravity and the radius of the circle since tension is 0. That means we can figure out velocity based on the relationship between gravity and the radius of the circle:

$v = \sqrt{rg} \equiv v = \sqrt{2 m*9.8 m/s^2} \equiv v = 2.21 m/s$

Now that we have the velocity and the time that it takes for the object to get to the ground we use the very simple formula d = vt in order to find the distance the object travels on the x-axis.

$d = vt \equiv d = (2.21 m/s)(0.64 s) \equiv d = 1.41 m$

Ta-da! Hopefully this is somewhat helpful in explaining how to get this answer. Please feel free to comment and or correct me if you find errors.

Caclulating distance related to tangential velocity

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