Introduction to Motion and Acceleration

Example 2 an object moves on a helical path given by

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: er is a vector; the latter a non-negative real- valued function. Example 2 An object moves on a helical path given by velocity and acceleration vectors and (2) the speed of the object at Solution (1) 4 , cos , sin 4, sin , cos 4, 0, sin , cos cos , sin 4 The speed is given by the norm of the velocity vector: sin 4 Solution (2) √17 3 cos 4 , cos , sin . Find (1) the 0. The speed of the object at 0 is 0 √17. Interestingly, this trajectory is one of constant speed. Can we therefore say that the velocity is constant? Example 3 An object moves on a path given by sin 1 cos . Sketch its trajectory. On the trajectory, plot the velocity and acceleration vectors at the point ( , 2 . Solution sin 1 cos /2 0 0 3 2 1 2 0 3 /2 1 2 2 1 1 0 The trajectory passes through the point ( , 2 when , that is , 2 , as can be seen from the table above. In order to sketch the trajectory, we may attempt to eliminate the parameter which does not look promising, or graph the functions and individually. Regardless of what we do, it is clear that tends to infinity although it oscillates in the process. However, since |cos | 1, 0 1 cos 2. Therefore, is a bounded function which oscillates between 0 and 2. The graph below illustrates the trajectory on the interval 0 4: ,2 2.0 1.5 1.0 0.5 0 2 4 6 8 The velocity and acceleration vectors at 2 12 are: sin 1 10 1 cos cos sin 0 Similarly, sin cos 0 These vectors are vectors in standard form so technically they have initial points at the origin. However, it is customary when dealing with motion to sketch them with their initial points at the point on the curve: 4 ,2 2 2.0 1.5 1.0 0.5 1 2 3 4 5 6 More often than not, when we study motion problems we know the forces at work, not the position vector. The problem is then to reconstruct the position vector from the acceleration vector. We do this via integration: Given ; 0 , , , , 0 , , , , , We use the dummy variable because it is good practice in mathematics not to use the same variable in an equation for two different purposes. 1 2 with the initial conditions Example 4 Given the acceleration vector 2 0 0,0 , 0 1,2 , determine the velocity and position vectors. What is the object’s speed at 1? Solution 0,0 1 2 | 2 | 0 0 1 1 1,2 1,2 3 1,2 1 3 2 3 The object’s speed at 2 2 2 1: 1 5 3 3 3 2 1 Therefore, 1 4 2 16 4 5.85 The units of...
View Full Document

This note was uploaded on 10/24/2012 for the course MAC 2313 taught by Professor Lopez during the Spring '10 term at Miami Dade College, Miami.

Ask a homework question - tutors are online