# Angular position vs. time graph

## Homework Statement

i am having trouble converting an angular v vs. time to position vs. time graph. i know that the slope of the position is the velocity but the velocity graph has line going up and down.

## The Attempt at a Solution

v v.s t = slope of p vs. t

#### Attachments

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Delphi51
Homework Helper
ω = v/r so your ω vs t graph is a v/r vs t graph.
A unit of area under it represents vt/r or distance/r
Thus the area needs to be multiplied by r to get the distance (position - initial position).

ω = v/r so your ω vs t graph is a v/r vs t graph.
A unit of area under it represents vt/r or distance/r
Thus the area needs to be multiplied by r to get the distance (position - initial position).
i am still confused...i have attached a sample...how would you draw the angular position vs. time and angular velocity vs. time.

Delphi51
Homework Helper
Oh, that is completely different from what I thought you asked!
No conversion from rotational to linear.
To go from angular acceleration to angular velocity, simply do the area under the graph. So in the first 0.5 s, the area is 0.4*0.5 = 0.2. On your ω vs t graph, you now have a point (0,0) assuming it starts with initial ω = 0, and a point (0.5,0.2). During this time the acceleration was constant so the ω should increase linearly (as in ω = α*t) so you can join the two points with a straight line.

When you do the next half second, note that the area is negative (below the zero line) and that this negative area must be added to the previous area (0.2 - 0.2 = 0) so your total area to time 1 second is zero. Third point(1,0).
The usual technique is to write in the areas for each half second and then sum them to get the points on the angular velocity graph.

Oh, that is completely different from what I thought you asked!
No conversion from rotational to linear.
To go from angular acceleration to angular velocity, simply do the area under the graph. So in the first 0.5 s, the area is 0.4*0.5 = 0.2. On your ω vs t graph, you now have a point (0,0) assuming it starts with initial ω = 0, and a point (0.5,0.2). During this time the acceleration was constant so the ω should increase linearly (as in ω = α*t) so you can join the two points with a straight line.

When you do the next half second, note that the area is negative (below the zero line) and that this negative area must be added to the previous area (0.2 - 0.2 = 0) so your total area to time 1 second is zero. Third point(1,0).
The usual technique is to write in the areas for each half second and then sum them to get the points on the angular velocity graph.
thanks that really did help...nbut how about angular position vs. time

Delphi51
Homework Helper
Take the area under the angular velocity graph to get angular position.
The areas will be a bit harder - area of a triangle is half the base times the height.
For accelerated motion, the angular position will increase parabolically so you'll have curved sections on the graph.

Take the area under the angular velocity graph to get angular position.
The areas will be a bit harder - area of a triangle is half the base times the height.
For accelerated motion, the angular position will increase parabolically so you'll have curved sections on the graph.
ok..thanks alot...iit really helped