Relations for 2-D motion
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Relations for 2-D motion. Motions that occur in 2-dimensions are more complicated since now the direction of the acceleration, velocity, force, etc. need to be taken into account.

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Relations for 2-D motion

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Relations for 2 d motion

Relations for 2-D motion

Motions that occur in 2-dimensions are more complicated since now the direction of the acceleration, velocity, force, etc. need to be taken into account.

These vectors, , and , are still proportional to each other by the proportionality constant, mass, and the two vectors are in the same direction.

Another important difference is that instead of taking time derivatives of a single valued function x, (i.e. a scalar), we now have to take successive time derivatives of a vector .


How do we take time derivatives of a vector

How do we take time derivatives of a vector?

The answer is that it depends….. It depends on which coordinate system we choose as convenient for a specific problem. For 2-D motion, we will work in Cartesian and Polar coordinates. First, let’s consider how to write functions and take derivatives in Cartesian coordinates

FORCE

ACCELERATION

Are unit vectors in the x- and y- directions (i.e. having magnitude of 1 and whose purpose is only to indicate direction)


Time derivatives in cartesian coordinates

Time derivatives in Cartesian Coordinates

Why is it so convenient to work in Cartesian coordinates? The answer is that the unit vectors do not change in time. They always point in the same direction.

Consider the vector:

Thus, if the force is only in the x- or y- direction, then the trajectory of the object can be computed the same as for 1-dimensional motion.


Polar coordinates

Polar Coordinates

In certain types of motion, such as revolution about an axis, it is sometimes more convenient to use polar coordinates. The main difficulty in polar coordinates is that the unit vectors can change direction as a function of position or time.

The transformation between Cartesian and Polar unit vectors can be written as follows:


Converting back to cartesian from polar coordinates

What is ?

Converting back to Cartesian from Polar Coordinates

Given

The transformation between Polar and Cartesian unit vectors can be written as follows:

Given these forward and reverse transformations between Cartesian and polar coordinates, the next question is how to take time derivatives of vectors written in polar coordinates? Suppose there is a vector, written as follows:


Time derivatives in polar coordinates i

Time derivatives in Polar Coordinates: I

Hence the velocity in polar coordinates can be given in terms of an object’s radial position, radial velocity, and rotation frequency.


Time derivatives in polar coordinates ii

Time derivatives in Polar Coordinates: II

This is a very interesting result, but is it useful? Why can’t we solve everything in the much simpler Cartesian coordinates? It turns out that a certain class of forces, known as “central forces” are most easily described using polar coordinates.


Relations for 2 d motion

Solution


Relations for 2 d motion

where

With Initial Conditions


Relations for 2 d motion

When is a maximum?

Answer: It is maximal when

Given that we know the y position of the projectile for t=0 and t=final are at the same position, we can solve for t.

In other words, at this angle the projectile can travel a maximal distance. So, what initial velocity is required for the projectile to travel 12km?


Relations for 2 d motion

Initial Conditions:

Terms we don’t know


Relations for 2 d motion

Solving for velocity magnitude

Solving for acceleration magnitude


Relations for 2 d motion

Problem 1

As observed from a ship moving due east at 8 km/h, the

wind appears to blow from the south. After the ship has

changed course, and as it is moving due north at 8 km/h,

the wind appears to blow from the southwest. Assuming

that the wind velocity is constant during the period of

observation, determine the magnitude and direction of the

true wind velocity.


Relations for 2 d motion

Problem 1

As observed from a ship moving due east at 8 km/h, the

wind appears to blow from the south. After the ship has

changed course, and as it is moving due north at 8 km/h,

the wind appears to blow from the southwest. Assuming

that the wind velocity is constant during the period of

observation, determine the magnitude and direction of the

true wind velocity.

1. To solve relative motion problems:

a. Write the vector equation that relates the velocities of two

particles.

vB = vA + vB/A

vB and vA are the velocities of particles B and A relative to a fixed

frame, respectively. vB/A is the velocity of B relative to A.


Relations for 2 d motion

Problem 1

As observed from a ship moving due east at 8 km/h, the

wind appears to blow from the south. After the ship has

changed course, and as it is moving due north at 8 km/h,

the wind appears to blow from the southwest. Assuming

that the wind velocity is constant during the period of

observation, determine the magnitude and direction of the

true wind velocity.

b. Express all velocity vectors in terms of their rectangular

components.

c. Scalar equations. Substitutethe velocity vectors in

vB = vA + vB/Aand solve the two independent sets of scalar

equations obtained.


Relations for 2 d motion

Problem 1 Solution

N

y

vs1

vw/s1

E

x

vs1 = 8 i km/h

vw/s1 = vw/s1j

vw = vwxi + vwyj

First observation

Write the vector equation

that relates the velocities

of two particles.

vw = vs1 + vw/s1

Express all velocity vectors in terms

of their rectangular components.

vwx = 8 + 0

vwx = 8 km/h

Scalarequations.

vwy = 0 + vw/s1


Relations for 2 d motion

Problem 1 Solution

Second observation

Write the vector equation

that relates the velocities

of two particles.

N

y

vs2

vw/s2

vw = vs2 + vw/s2

45o

Express all velocity vectors in terms

of their rectangular components.

E

vs2 = 8 j km/h

x

vw = vwxi + vwyj

vw/s2 = vw/s2 cos 45oi + vw/s2 sin 45oj

Scalar equations.

vw = 8 i +16 j km/h

vwx = 0 + vw/s2 cos 45o

vwy = 8 + vw/s2 sin 45o

N

vw = 17.89 km/h

63.4o

vwy = 16 km/h

E


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