Angular Velocity and Acceleration graph

Click For 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
Messages
27
Reaction score
1

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 :(
 
Physics news on Phys.org
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?
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat (moth as source and bat as observer): $$f_2=\frac{333}{333-v}\times f_1$$ $$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$ Solving this equation, I get ##v=6.1## m/s but the answer key is...