Motion in Two Dimensions Lecture Notes PDF

Motion in Two Dimensions By Dr/Mai Hosny Nasr Dr/ Mai Hosny Nasr Vector Components From the definitions of sine and cosine, the rectangular components of 𝑨 x, namely Ax and Ay, will be given by: Dr/ Mai Hosny Nasr  Projectile Motio...

Motion in Two Dimensions By Dr/Mai Hosny Nasr Dr/ Mai Hosny Nasr Vector Components From the definitions of sine and cosine, the rectangular components of 𝑨 x, namely Ax and Ay, will be given by: Dr/ Mai Hosny Nasr  Projectile Motion  Any object that is thrown into the air is called a projectile. Near the Earth’s surface, we assume that the downward acceleration due to gravity is constant and the effect of air resistance is negligible.  Based on these two assumptions, we find that: (1) the horizontal motion and vertical motion are independent of each other and (2) the two dimensional path of a projectile (also called its trajectory) is always a parabola. Dr/ Mai Hosny Nasr Dr/ Mai Hosny Nasr The components vx◦ and vy◦ can then be found in terms of the initial speed v◦ and the launch angle θ◦ as follows: Horizontal Motion of a Projectile the horizontal velocity component vx◦ remains constant throughout the motion, Thus, the horizontal velocity vx and the horizontal position x are described as follows Dr/ Mai Hosny Nasr Vertical Motion of a Projectile the vertical motion is the motion we discussed for a particle in free fall. Thus, the vertical velocity vy and the vertical position y can be described as follows: Dr/ Mai Hosny Nasr Horizontal Range of a Projectile The horizontal range R is the distance traveled by the projectile when it returns to y = 0 after time t = T, as seen in the previous figure. To find an expression for R we use the pervious equations. Thus: From the last result, we get the following relation for T: Dr/ Mai Hosny Nasr Example 1: An airplane is flying horizontally with a constant speed v◦ = 400 km/h at a constant elevation h = 2 km above the ground, see Fig. (a) If the pilot decided to release a package of supplies very close to a truck on the ground, then what is the time of flight of the package? (b) What is the horizontal distance covered by the package in that time (which is the same horizontal distance covered by the plane)? Solution: (a) The initial velocity of the package is the same as the velocity of the plane. Therefore, the initial velocity of the package →v ◦ is horizontal (i.e., θ◦ = 0) and has a magnitude of 400 km/h Dr/ Mai Hosny Nasr the vertical distance that the package falls, then we find its time of flight from Eq. as follows: (b)The horizontal distance covered by the package in that time is: Dr/ Mai Hosny Nasr Example 2: A basketball player throws a ball at an angle θ◦ = 60◦ above the horizontal, as shown in the Fig. At what speed must he throw the ball to score? Solution Thus, solving for v◦ and taking the positive root gives: Dr/ Mai Hosny Nasr Uniform Circular Motion  A particle that moves around in a circle with a constant speed, like the car shown, is said to experience a uniform circular motion. In this case, the acceleration arises only from the change in the direction of the velocity vector.  the acceleration associated with uniform circular motion is called centripetal acceleration Dr/ Mai Hosny Nasr  the particle travels the circumference of the circle in a time T giving by: where T is called the period of revolution, or simply the period. Example1: A satellite is circulating the Earth at an altitude h = 150 km above its surface, where the free fall acceleration g is 9.4 m/s2. The Earth’s radius is 6.4 × 106 m. What is the orbital Speed and period of the satellite? Dr/ Mai Hosny Nasr As shown in Fig, the radius of the satellite’s circular motion equals the sum of the Earth’s radius R and the altitude h, i.e. By using the centripetal acceleration Eq, we find that the magnitude of the satellite’s acceleration can be written as: For the uniform circular motion of the satellite around the Earth, the satellite’s centripetal acceleration is then equal to the free fall acceleration g at this altitude. That is: Dr/ Mai Hosny Nasr  From the preceding two equations we have: Solving for v and taking the positive root gives With this high speed, the satellite would take T = 2πr/v = 1.46 h to make one complete revolution around the Earth. Dr/ Mai Hosny Nasr