Tuesday, January 26, 2010

Circular Motion and Gravitation

Although grasping the concept of circular motion and gravitation can be difficult, I learned many new things about physics. When an object is in a constant or uniform speed traveling in a circle, the object is in uniform circular motion. Also, the distance and object travels around a circle is the perimeter. One complete revolution around the perimeter of a circle is the circumference. The equation 2*pie*r represents the circumference of a circle. In addition, the speed of the object in uniform circular motion is given by v=2*pie*/T m/s. Period T is the variable for the time it takes for the object to make one full revolution around the perimeter, and it is given in seconds. Also, the frequency is the number of rotations per unit of time an object makes around the perimeter, and is given in Hertz, abbreviated as Hz. Therefore, T=1/f s and f=1/T Hz. Also, objects moving at a constant speed don't have a constant velocity. This is because an object in uniform circular motion is constantly changing direction. However, the magnitude of the velocity of the object will remain constant. Tangential is best used to describe the direction of the velocity vector of an object while in uniform circular motion. Furthermore, when an object changes direction, it accelerates. This acceleration is called centripetal acceleration. Centripetal acceleration means that the acceleration will always be directed towards the center of the circle.  Another fact is that the acceleration will always be perpendicular to the velocity. In addition, centripetal force is the force that must be applied to keep an object moving in a circle. The centripetal force is not a force by itself, but the centripetal force is provided by the force that keeps the object in a circle. the equation to solve for the centripetal force is Fc=mv^2/r N. When an object moves in a vertical circle at the end of a string, the tension varies with the position of the body. At the highest point the centripetal force equals Ft+mg=mv^2/r N. When it is at the lowest point Fc is found by Ft-mg=mv^2/r N. We also learned that Newton discovered that the gravitational force varies inversely with the square of the distance between two objects, which is called the inverse square law.  The Law of Universal Gravitation states that "Every object in the universe attracts every other object in the universe with a force that varies directly with the product of their masses and inversely with the square of the distance between the centers of the two masses." Therefore the equation for the force of gravitation is Fg=Gm1m2/r^2 N. To find the acceleration due to gravity of an object of mass we can use Newton's Law of Universal Gravitation. The equation can be manipulated into another equation: g=GM/r^2. We have learned a lot this unit, but overall it has been very interesting.

What I have found difficult about what we have learned is the period and frequency. I constantly get confused and mixed up between the both. Also, I find it difficult to find the centripetal force requirement, especially when many forces are acting on the object.However, I feel that I can get confident in these areas of weakness if I keep practicing.

My problem solving skills are pretty good I would say. Sometimes I can make careless mistakes and not realize it. Other times the answer is right in front of me, but I just can't see it. The key to my success is practice because that is how I learn. It gives me more and more experience and prepares me to solve even harder problems in that area. If I continue to do this, I believe I can get better at problem solving.

Sunday, January 10, 2010

Newton's Second Law

This is what I learned about Newton's second law...

Newton's second law states that the acceleration of an object is directly proportional to the net force, and indirectly proportional to the mass of the object. When I first saw this, I have to admit that I was very confused. However when I was given the equation F=ma, I understood the concept. "F" is the variable for the sum of the forces, "m" stands for mass, and "a" stands for acceleration. This equation is the foundation for a=F/m. The second equation is used to find the acceleration of an object that is under contact of a force. Now I understand this law thoroughly. An easy way for me to think about it is if someone pushed a ball across the floor with an applied force, then the object would obviously accelerate if you think about it. Also, if you pushed it four times as hard, then the ball would accelerate four times as fast. This thought made the law very clear to me. During this unit, we also learned about apparent weight. Apparent weight is the amount of force an object exerts on the surface it is on, where as actual weight is the amount of gravitational force that acts on an object. Another thing we learned was about friction."The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it." There are two types of friction, kinetic and static. Kinetic friction occurs when an object is in motion, and static friction occurs when an object is at rest. We also learned about the coefficient of friction,mu. The equation is Ff=Fn*mu. We have learned a lot about this unit, and so far it has been filled with fun!

What I have found difficult about what I have studied is when to set the sum of the forces equal to ma. At first, I was very confused about this, but I later figured it out: the sum of the forces is set to ma if the axis that the motion is occurring in is being involved in the sum of the forces equation, and the sum of the forces is equal to zero if the axis where the motion is not occurring is involved in the sum of the forces equation. I also was a little confused about when to find the sum of all the forces and when to find the sum of only the forces in the axis of motion. These two things were the most confusing for me, but it didn't take me long to figure them out.

I believe that my problem solving skills have improved tremendously. I am very efficient now, and I am able to tell how I can figure out various unknown variables in problems. I am also very comfortable with the FBDs and the new equations. Overall, I believe that I have gained more confidence in being able to solve problems.