A physics problem in my book that I thought was hard

A particle starting from rest slides down a frictionless ramp. It is to attain a given horizontal displacement in a minimum amount of time. What is the best angle for the ramp? What is the minimum time?

Usually I'll solve something and realize that it wasn't that bad, but I still think this is hard...sort of.

Start out by drawing a diagram of the particle sliding down a ramp or incline. This is just to give an idea of what we're dealing with. The arrow I've drawn indicates the direction of the particle's motion.

Draw another diagram of the particle that shows its acceleration vectors. The vector pointing straight down is the acceleration due to gravity (g). The other two are components of g, one of which is in the direction of the incline and the other of which is perpendicular to the incline. We will focus on the acceleration vector that points in the direction of the incline because this is the one that causes the particle to accelerate down the plane.

As the particle moves down the plane, it experiences a displacement in the direction of the ramp denoted by delta z; however, we are interested in the horizontal displacement. We will break up the displacement vector delta z into a horizontal and vertical component denoted by delta x and delta z respectively. Now we should be able to write an equation of the horizontal displacement in terms of delta z and theta.

Keep this equation in mind because we will use it later. Right now we will focus on writing the displacement of the particle in the direction of the plane (or in the direction of delta z) in terms of the change in time and the angle theta.

Now this is where the equation above comes in: Once we figure out and equation for the displacement in the delta z direction , we will multiply the whole thing by cos(theta) to obtain an expression of the displacement in the horizontal (delta x) direction.

Also Recall from the acceleration diagram that the acceleration in the direction of the plane is equal to g*sin(theta). This is what we'll use in our kinematics equation.

we have now written the horizontal displacement in terms of theta and the change in time.

Want we want to know is what angle of inclination will cause a certain horizontal displacement in a minimum amount of time. In other words, for what angle does the particle experience the greatest displacement?

It may have struck you that this problem becomes similar to finding the relative maximum of a function. It almost is. The tricky thing here is that the equation is in terms of two variables, not one.

In this case when we solve for the relative maximum of the equation, we will treat the change in time as a constant and differentiate the horizontal displacement with respect to the angle theta. If you think about it, the change in time is really irrelevant in this situation because it's not what causes a change in the vector components of the particle's acceleration. We are interested in the variable that causes a direct change in the particle's acceleration components, so we differentiate with respect to theta.

Now that we have the angle that will cause a horizontal displacement in a minimum amount of time, we simply plug it back into the original equation (the one we differentiated) and solve for delta t.

Solution to the physics problem I found

You may have noticed that as object A moves to the left there are several variables that change with respect to time. The vertical height y increases, the horizontal length x decreases, and the angle alpha increases. This is really no different from an ordinary related rates problem, so we'll treat it like one.

Start out by using the Pythagorean theorem to write the vertical length y as a function of the horizontal length x.

recall that the derivative of displacement with respect to time is equal to velocity. So in order to create a velocity equation, we will differentiate both sides of the equation above with respect to time.

We're almost ready to solve the problem. All we have to do now is a little bit of substitution and manipulation of the problem.

important note* dx/dt is the velocity of A. It's important to know that the velocity of A is going to be negative because as A is moving to the left, the length of side x gets smaller. This means that the rate of change of x with respect to time (dx/dt) is negative. It is thus substituted with negative v.

at this point it's just a matter of rearranging parts of the equation and making it look nice. It should simplify to something fairly simple where we can plug in the angle alpha.

Isn't it mind blowing? We started out with a nasty looking velocity equation and simplified to something simple.

A cool physics problem I found

I found this problem in my physics book. It really stumped me for a while, but I finally figured it out after some time. It looked intimidating, but it's not as bad as I thought. I tried to redraw the diagram as closely as I could and I almost copied the problem down word for word.

Two objects, A and B, are connected by a rigid rod that has a length L. The objects slide along perpendicular guide rails, as shown in the figure. If A slides to the left with a constant speed v, find the velocity of B when the angle, alpha equals 60 degrees.

Solution to the cool integration by parts problem

This problem requires a bit more ingenuity.

Start out using integration by parts as you normally would.

You may notice that the new integral acquired still poses the same issue we started out (but with cosine instead of sine). Let us do integration by parts once more and see what happens

Now I need to substitute this back in. I need to also make sure to distribute the negative.

Now we have the exact same integral we started out with on both side. all we really need to do is add the integral of e^(x)sin(x) to both sides of the equation. Then we'll have 2 times the integral of e^(x)sin(x) on one side. Simply divide by two and the whole thing will be solved.

Is your mind blown or what?

A cool integration by parts problem

Here's a cool integral that requires integration by parts. It's a little tricky, but the solution is mind blowing.

Deriving the volume of a sphere solution

note* I'm assuming you are already familiar with calculus. If you're not, it's probably not going to make any bit of sense.

I will be doing this using integration techniques. Remember learning about solids of revolution from calc. I? That's what I'll be using. I'm not really sure if it's the best way, but it works.

Consider the graph of a basic semi-circle. If you were to revolve it around the x-axis, the solid generated would be a sphere whose cross sections are circles.

Imagine slicing through a sphere. You'll notice that the cross sections are circles of varying sizes. What I'll essentially be doing is creating a summation of disks that each have a very small thickness of dx.

To start off, I'm going to create a function of the cross sectional area and integrate it from one end of the semi circle (-r) to the other (r). I'll need the formula for the area of a circle. The radius of each cross section will vary as a function of x, and so I need to make the appropriate substitution for the radius. Through integration, the disks of varying cross sectional area will be summed together to create a sphere.

There's nothing all that tricky about solving this integral. If you work it all out you should end up with the correct formula for the volume of a sphere.

Deriving the volume of a sphere

For this next one I want to derive the volume of a sphere using calculus. I don't have much more to add, but I feel like I should at least post a picture of something...