Angular Velocity and Position: How to Solve Basic Angular Motion Problems

[Aristotle]

Homework Statement

Homework Equations

ω=ω0+αt

θ=θ0+ω0t+12αt2 I think?

The Attempt at a Solution

For part a, I know that from the graph, the angular velocity in this case would be constant since the graph illustrates a position in rad vs time graph. The thing is, I'm really sure on how to approach this problem. Could somebody hint me on how to work this problem?

 

[Aristotle]

For a, aren't we just subtracting the final angle and initial?

 

[BvU]

For a, aren't we just subtracting the final angle and initial?

Yep !

Your relevant equations are applicable for uniformly accelerated angular motion ($\alpha$ constant). Here, we clearly have something else at hand, so you will have to fall back to the definitions
($\omega = \dot \phi = {d\phi\over dt}$ and ($\alpha = \dot \omega = {d\omega\over dt}$ )
[edit] -- however, you don't need these for this exercise. Basically it's a math exercise :)

 

[Aristotle]

ω˙=dωdt\alpha = \dot \omega = {d\omega\over dt} )

Yep !

Your relevant equations are applicable for uniformly accelerated angular motion ($\alpha$ constant). Here, we clearly have something else at hand, so you will have to fall back to the definitions
($\omega = \dot \phi = {d\phi\over dt}$ and ($\alpha = \dot \omega = {d\omega\over dt}$ )
[edit] -- however, you don't need these for this exercise. Basically it's a math exercise :)

Ah I see. Oh and for angular displacement, do you typically consider them as vectors? Not sure if I should include the notation for thetha final - thetha initial...

 

[BvU]

Angles are scalars, not vectors. $\theta$ is in radians. It's a bit strange to put "Position in radians" on a graph (but understandable: angular displacement in radians is a mouthful).$\Delta \theta$ already has some sense of direction. It comes close to being a vector.

and from that $\omega$ and $\alpha$ are pseudo vectors (also this link, and this).

Easier to follow when you know about cross products and polar coordinates. Is that familiar already ?

 

[Aristotle]

Angles are scalars, not vectors. $\theta$ is in radians. It's a bit strange to put "Position in radians" on a graph (but understandable: angular displacement in radians is a mouthful).$\Delta \theta$ already has some sense of direction. It comes close to being a vector.

and from that $\omega$ and $\alpha$ are pseudo vectors (also this link, and this).

Easier to follow when you know about cross products and polar coordinates. Is that familiar already ?

Yup! Just found about that, much thanks!
Last question was for Part B, do you think I'm on the right track in finding the exponential function of the log-log graph scale (e.g: slope of line of log-log: n= logx2-logx1/logt2-logt1 andetc) to find the equation of the angular position of the object?

 

[BvU]

For part b) you indeed make good use of the observation that the relationship between log(theta) and log(time) is linear. I understand what you mean, but to make it unambiguous, better use some brackets: n= (logx2-logx1)/(logt2-logt1) :)

For c) you use the expression and check with the plot. Don't answer in five digits: the accuracy allows two, max three.

 

[Aristotle]

For part b) you indeed make good use of the observation that the relationship between log(theta) and log(time) is linear. I understand what you mean, but to make it unambiguous, better use some brackets: n= (logx2-logx1)/(logt2-logt1) :)

For c) you use the expression and check with the plot. Don't answer in five digits: the accuracy allows two, max three.

That's the problem, I tried finding the equation for the log-log plot, however ended up getting Ꝋ(t)=(3 rad/s^2) t^2. Whenever I use some arbitrary point of time from the graph, I don't get a corresponding result that matches up with the y-axis (position in rad).

The way I solved this was n= log(200,000)-log(3) / (log 42)-log(11) = 2.972
Then knowing that the position function is parabolic in shape: x(t)=Ct^2. So we need to find proportionality constant we solve log x = log C + 3 log(t). So to make it log(x)=log(c), I use t=1s (because log1=0), and got log(3)=logC + 0 ==> C=3.

So once I found 3, i then get the answer Ꝋ(t)= (3)t^2...am I doing something wrong here?

 

[BvU]

knowing that the position function is parabolic in shape: x(t)=Ct^2.

is not right. That is only for constant angular acceleration, which is clearly not the case here.

You found a linear relationship $\ln(\theta) - ln(3) = A \;( ln (t) - ln(1) )$ with A = 2.972 as the slope of the line. So now they want you to rewrite $\theta =$ ?

 

[Aristotle]

So now they want you to rewrite θ=\theta = ?

Oh okay, it works with
θ(t) = (3 rad/s^2) t^(2.972). In some sense correct now?

 

[BvU]

In a physics sense: excellent. And:
Easy to check: fill in a few t and look at the picture !

But (math,nitpicking, ...) : I would crank up the 2.972 a little bit. root cause: your 200000 looks to me already achieved at t = 40.

In part c) notice how small changes in the slope and the intercept give big changes in theta. That's what you get for using log-log plots...

 

[Aristotle]

In a physics sense: excellent. And:
Easy to check: fill in a few t and look at the picture !

But (math,nitpicking, ...) : I would crank up the 2.972 a little bit. root cause: your 200000 looks to me already achieved at t = 40.

In part c) notice how small changes in the slope and the intercept give big changes in theta. That's what you get for using log-log plots...

Yeah I rounded the 2.972 to 3.
For c) I just took the derivative of Θ(t) to get w(t)= 138 rad/s

 

[BvU]

Must be an empty battery in the calculator.
You can clearly see in the figure that $\theta$ increases a lot more every
second than that at t=23.
How did you calculate this 138 rad/s ?

 

[Aristotle]

Must be an empty battery in the calculator.
You can clearly see in the figure that $\theta$ increases a lot more every
second than that at t=23.
How did you calculate this 138 rad/s ?

Derivative of Θ(t) and got w(t)=6(t) and so for 23 seconds, got 138 rad/s

 

[BvU]

Derivative of which expression for $\theta$ ? The wrong one, or the one you found in b) ? what is the derivative (not the number, the expression) ?

 

[Aristotle]

Derivative of which expression for $\theta$ ? The wrong one, or the one you found in b) ? what is the derivative (not the number, the expression) ?

the one I found for b.

 

[Aristotle]

apologies what I meant was w(t)=9t^2 which I got from theta(t)=3t^3. :)

 

[Aristotle]

Correct now?

 

[BvU]

Yep :)
(around 4800 rad/s, I hope ?)

 

[Aristotle]

Yep :)
(around 4800 rad/s, I hope ?)

Yup what I got! I really appreciate your help bvu :D