Dynamics and relativity 1

Question 1

positions, velocity and acceleration.

Answer 1

Position is the position vector R from origin. described as xi + yj + zk. Velocity and acceleration are derivatives of position. all are functions of time. r can be expressed in terms of t using taylor expansion

Question 2

One dimensional motion

Answer 2

only considering one dimension, we can progressively integrate from acceleration (assuming constant a) to reach an expression for x (or any other) in terms of t. This is the same as the one directional version of the taylor series

Question 3

Freely falling motion

Answer 3

using one dimensional expression assuming constant acceleration is -g and there is no resistance.

Question 4

projectile motion

Answer 4

combining one dimensional expressions for x and y to track a projectile. These expressions can be used independently, or combined to form the equation of a parabola. Roots of parabola are equal to range, one is zero the other is range. taking x is half of the range will give max height, as parabolas are symmetric.

Question 5

Relative motion

Answer 5

depending on frame, we cannot agree on origin, we write A/B as the point a from the perspective of B. taking a point p, and two perspectives a and b then: p from a = p from b + b from a

Question 6

circles

Answer 6

2 dimensional motian in terms of r and theta. new symbols are angular velocity and angular acceleration in terms of theta instead of distance. tangential acceleration a tracks force towards centre of circle.

Question 7

newtons laws

Answer 7

take all forces in x and y and find resultant forces (first law), considering normal reaction forces (third law). apply F=ma to find motian (second law) use force diagrams.

Question 8

SHM

Answer 8

we define a constant restoring force towards equilibrium position F = -kx. where x is displacement from equilibrium. Then we define A, T, F and w. using diff eq we get the equation x = Acos(wt + phi) and differentiating v =-Awsin(wt + pjo)

Question 9

simple pendulum

Answer 9

F = -mgsintheta. using small angle approximation k = -mgtheta where theta = x/L. so for shm to take place omega must be equal to sqrt(k/m) which in this case simplifies to sqrt(g/L)