The next unit we studied this year was circular motion. Although I had learned some of the more simple concepts of this unit in grade 11, there were many new, challenging topics to learn. To begin with, we learned about uniform circular motion, which is when an object is moving with a constant speed in a circle, however it must be noted that the direction of motion is always changing, so the object is also accelerating. A demonstration in class showed that the direction of velocity is always tangent to the radius of the circular path, when a marble was swirled in a circular inside a roll of tape. When the tape was lifted and the marble continued to roll, it went straight in the direction it was moving without the circular path. This means that an object moving in a circle is accelerating toward the center of the path, and this is called centripetal acceleration.
When calculating the magnitude of centripetal acceleration (Ac) it is important to notice that the quantity is scalar because the direction of Ac is always towards the middle of the circle. The equation used for calculating this is Ac=v^2/r, where v is velocity, and r is the radius of the circle. If we are trying to find the total distance covered to complete one rotation of the circular path, we can find the circumference, by using v=2πr/T, where T is the period in seconds.
When solving problems with centripetal force (Fc), Fc is the sum of all the forces on an object moving in a circle, which means that it can include forces such as gravitational force or friction. Therefore, when we solve for Fc, we can expand to use Ac formula, leaving Fc=mAc=m(v^2/r).
Because many of the problems we do involve the Ac of cars turning corners, we also learned a practical application of this when we learned about banked curves. When you draw a free body diagram for a car turning a regular corner, we can see that friction is the centripetal force that keeps the car on the track. However, banked curves are when the surface itself is at an angle. This is much safer because it does not rely on friction to keep the car on the track. This means that cars can go faster on banked curves and be safe, while car will have to go slower around regular corners to be safe. The formula we use for banked curves is v=√rgtanθ, because θ is the angle of the curve, and we are using the parallel component of Fg to solve. While car problem involve horizontal circles, we also learned about vertical circular motion.
http://www.maa.org/publications/periodicals/loci/joma/banked-curves
Many of the example we do for vertical circular motion involve Ferris Wheels, which I will discuss in the application post, but just to get a general sense, Ac is still always in the middle of the circle, however, when an object is at the top of the circle, Fg and Fn are different than when the object is at the bottom of the circle.
The next part of circular motion that we learned was about gravitational circular motion, and orbital mechanics. From grade 11, we reviewed the Law of Universal Gravitation which is Fg=GMm/r^2, where G is the universal constant (6.67x10^-11). We also learned how to find the gravitational field strength that other planets exert on objects, such as Jupiter. To find this, we use g=GM/r^2.
A really important point to remember about this unit is that Fc=Fg when objects are in orbit. When we substitute formulas that we have already learned, we end up with m(v^2/r)=GMm/r^2, which leaves us with v=√GM/r. R in orbital mechanics,means the radius of the planet and the height at which the object is orbiting because the distance measured between object is from the center of each object.
The last section we learned was orbital mechanics. Orbital potential energy is relative, which means that the potential energy of an object changes depending on the reference point we are dealing with. To calculate potential energy (Ep) we use the formula, Ep=GMm/r. It is important to remember that for the potential energy of objects in orbit, the phrase "relative to zero at infinity" is used, which means that at an infinite distance from Earth means that the object's potential energy is 0. Therefore, when an object is closer to Earth, the potential energy is negative. When trying to find the amount of work done, we go back and use the conservation of energy, which is W=
ΔKE+
ΔPE. Escape velocity is when an object escapes the pull of gravity, and essentially travels an infinite distance from the planet. For this, we also use the conservation of energy, leaving Vesc=√2GM/r.
http://physics.nayland.school.nz/VisualPhysics/NZ-physics%20HTML/06_Gravitation/chapter6d.html
With this unit, the aspect that I had the most difficulty with was trying to derive the formulas required to solve for different values. For many of the problems we did, asked to solve for the radius or period, which there is no explicit equation to use to solve for. Instead, we had to substitute different equations in and derive the formula needed. Although this does not seem overly difficult, it was an adjustment that took a lot of practice to get used to doing in a timely manner. Another concept that I struggled to remember was that the term "radius" meant that the distance measured is from the center of the 2 object, and includes the height that the object is orbiting. Although I understand this, I sometimes forgot to include the appropriate heights when doing calculations.
