Rotation Problem - Constant Acceleration

[FaraDazed]

Homework Statement

[/B]
At t=0, a flywheel has an angular velocity of 3.4 rad/s, an angular acceleration of -0.42 rad/s^2 and a reference line at \theta_0 =0

A: Throigh what maximum angle will the reference line turn in the positive direction

B: At what times will the reference line be at \theta_{max} /7

Homework Equations

\theta=\theta_0 + \omega t + \frac{1}{2}\alpha t^2

The Attempt at a Solution

This problem is doing my head in, so would appreciate a little help/advice/feedback please :)

For a, theta max will occur when the derivative of that equation in the relvevent equation section is zero.

<br /> \frac{d \theta}{dt}=\omega + \alpha t = 0 \\<br /> t=-\frac{\omega}{\alpha}<br />
Then subbing that into the original equation gives
<br /> \theta_{max}=-\frac{\omega^2}{\alpha} - \frac{1}{2}\frac{\omega^2}{\alpha} \\<br /> \theta_{max}=-\frac{3}{2}\frac{\omega^2}{\alpha} = -\frac{3}{2}\frac{3.4}{-0.42} = 41.29 \,\,\, rad/s<br />

Then for b i did
<br /> \frac{\theta_{max}}{7}=\omega t + \frac{1}{2} \alpha t^2 \\<br /> -\frac{\frac{3}{2}\frac{\omega^2}{\alpha}}{7} = \omega t + \frac{1}{2} \alpha t^2 \\<br /> -\frac{21 \omega}{2 \alpha} = \omega t + \frac{1}{2} \alpha t^2\\<br /> -21 \omega = 2 \alpha \omega t + \alpha^2 t^2 \\<br /> 0 = \alpha^2 t^2 +2 \alpha \omega t + 21 \omega<br />

But when I use the quadratic formula, I always get a math error on my calculator, because 4ac > b^2 . I think my method is sound (I can't think of any other way of doing it) but I must be making an algebraic mistake somewhere, so would appreciate a second pair of eyes.

 

[FaraDazed]

To make part B easier, now I just did

<br /> \frac{\theta_{max}}{7}=\omega t + \frac{1}{2} \alpha t^2 \\<br /> \frac{41.29}{7}=3.4t + \frac{1}{2} (-0.42) t^2 \\<br /> 0= \frac{1}{2} (-0.42) t^2 + 3.4t - 5.89<br />

Then using quadratic formula for first time
<br /> t_1 \frac{-3.4+\sqrt{ 3.4^2-(4 \times 0.5 \times (-0.42) \times -5.9)} } {2 \times 0.5 \times -0.42 } =1.97s<br />

And for second
<br /> t_2 \frac{-3.4-\sqrt{ 3.4^2-(4 \times 0.5 \times (-0.42) \times -5.9)} } {2 \times 0.5 \times -0.42 } =12.98s<br />

 

[FaraDazed]

Anyone? Really need help with this, I am not sure I did the expression correct for part a as I ha e seen it as q/2 rather than 3/2 on a peers work.

 

[nasu]

The time is not negative. The angular acceleration is.
So when you solve for time, in part (a), you get positive time.
Then your value for maximum angle is not right, due to this sign error.
The first term (ω t) is positive.

 

[BvU]

1)
$\theta_{max}=-\frac{\omega^2}{\alpha} - \frac{1}{2}\frac{\omega^2}{\alpha}$ isn't right $\ \ \theta_{max}=-\frac{\omega^2}{\alpha} + \frac{1}{2}\frac{\omega^2}{\alpha}$ is...

2)
$\theta_{max} = ...\,\,\, rad/s$ isn't right $\ \ \theta_{max} = ... \,\,\, rad$ is. Looks like nitpicking, but things like that can cost you precious points (or lead you to the wrong expressions).

3) ${3/2 \over 7}$ is not $21\over 2$ (I take it the $\omega$ instead of $\omega^2$ is a typo...) And the 3 should be a 1 (see 1) )

4)
Since the thing is slowing down, I don't know anything better to do than solve for $\theta = {\theta_{max}\over 7 } + 2n\pi$ with n = 0, 1
Unfortunately, it then starts turning the other way around and passes the same angular position in ever more rapid successions...perhaps they don't want those, only the first two from the four you get from n = 0 and 1. (i.e. only the passes while still turning in the positive direction)

 

[nasu]

1)
$\theta_{max}=-\frac{\omega^2}{\alpha} - \frac{1}{2}\frac{\omega^2}{\alpha}$ isn't right $\ \ \theta_{max}=-\frac{\omega^2}{\alpha} + \frac{1}{2}\frac{\omega^2}{\alpha}$ is...

Should't the signs be the other way? It says that it turns in the positive direction. So the angle should be "positive". And the angular speed is positive, isn't it?

 

[BvU]

$\alpha<0$

 

[nasu]

Oh, you consider alpha with its sign. I did not realize this.

 

[FaraDazed]

Ok thanks for your help guys, the comments have helped. I have had another go and got the same result as two of my peers so think its correct.

I think the lesson to learn from this is not to work on a hard problem when really tired, lots of mistakes! :D