Angular Velocity and Acceleration graph

AI Thread Summary
The discussion focuses on determining angular displacement and acceleration from a given graph. For angular displacement during the 4 seconds of motion, the area under the graph is the correct method to calculate it. There is confusion regarding which angular velocity (ω) value to use for calculations, with participants debating whether it should be constant or varying. For angular acceleration between t=2 and t=4, it is clarified that the acceleration is constant, despite some initial uncertainty. Understanding the relationship between velocity and acceleration on a graph is emphasized as crucial for solving these problems.
df102015
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Homework Statement



04.EX32.jpg

A.) For the graph above what is the angular displacement during the 4 seconds of motion?
B.) For the graph above what is the angular acceleration from t=2 to t=4?

Homework Equations


α = at / r
α = ω / t
α = Θ / t^2
ω = Θ / t
ω = v / r
Θ = ω t + 0.5 α t^2

The Attempt at a Solution



A.) I used ω = Θ / t
rearranged it to ωt = Θ

but my issue is what ω do i use? 0, 10, 20? anything in between? And is this even the right equation to use?

B.) Isn't it not accelerating between t2 and t4? Or is it constant acceleration?
How i went about it is i used α = ω / t
since the time is 2 seconds, and the ω is 20. i got 10, but that was wrong :(
 
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There is an error in your image, unfortunately.
 
ProfuselyQuarky said:
There is an error in your image, unfortunately.
fixed it
 
For the first part it is the area under the graph.
For the second part the angular acceleration is constant.
 
df102015 said:
what ω do i use?
Do you know how to find the distance moved from a velocity-time graph?
df102015 said:
α = ω / t
Quoting formulae is of little use unless you know what the terms in the formulae represent. E.g. a=v/t is more informatively written as aavg=Δv/Δt. I.e. the average acceleration is the increase in velocity divided by the elapsed time.
kinemath said:
For the second part the angular acceleration is constant
True, but of the choice
df102015 said:
not accelerating between t2 and t4? Or is it constant acceleration?
that response could be misleading.
@df102015 , what do you look at on a velocity-time graph to deduce the acceleration?
 
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