What is the Correct Equation for Angular Acceleration in this Flywheel Problem?

[k_squared]

Homework Statement

"The flywheel rotates with angular velocity of w=0.005theta^2 rad/s. Determine the angular acceleration after it has rotated 20 revolutions.

Homework Equations

I thought the equations were all but self-evident using the problem description! (See below.)
alpha=dw/dt

The Attempt at a Solution

The angular velocity is merely a=0.01theta. I just plugged in 20*2pi for the acceleration at that time and was... wrong. Massively wrong, in fact. The book solutions agrees with me about my equations for w and a, but they actually multiple the acceleration and the velocity together!

The book defines alpha as dw/dtheta... but then, in this solution set, states that alpha=w(dw)(dtheta), which is not something I can make *any* sense out of. They then use this value of alpha to compute the acceleration over a rotation of (40pi), complete with theta cubed.

What happened here? I can't make heads or tails of it.

 

[TomHart]

The book defines alpha as dw/dtheta

Are you sure of this definition?

 

[hilbert2]

You need to use the chain rule of differentiation when applying $\frac{d}{dt}$ on $[\theta (t)]^2$.

 

[Delta2]

due to the chain rule of differentiation it is $\frac{dw}{dt}=\frac{dw}{d\theta}\frac{d\theta}{dt}$ and it is also $\frac{d\theta}{dt}=w$ so what your book says is correct.

 

[Buffu]

The book defines alpha as dw/dtheta... but then, in this solution set, states that alpha=w(dw)(dtheta), which is not something I can make *any* sense out of. They then use this value of alpha to compute the acceleration over a rotation of (40pi), complete with theta cubed.

your book is wrong, it is not $$\alpha = w\times dw \times d\theta$$

it is,
$$\alpha = {d \omega \over dt} = \underbrace{{d\omega \over d\theta } \times {d\theta \over dt}}_{\text{chain rule}} = ?$$

 

[haruspex]

due to the chain rule of differentiation it is $\frac{dw}{dt}=\frac{dw}{d\theta}\frac{d\theta}{dt}$ and it is also $\frac{d\theta}{dt}=w$ so what your book says is correct.

... assuming k² meant alpha=w(dw)/(dtheta) and not

alpha=w(dw)(dtheta),

 

[Delta2]

guys I believe the post has some typos, I believe he meant to write that his book says that a=wdw/dtheta , $a=w\frac{dw}{d\theta}$ in latex :).

 

[k_squared]

I'll give you a screenshot if you want!

The book defined alpha (angular acceleration, I take it) merely as $$\frac{dw}{d\theta}$$. Then is also says that $$\alpha d\theta = \omega d\omega$$.

The exact answer has the line in it:
$$\alpha=\omega \frac{d\omega}{d\theta}$$ and using the given (.005theta) value and its derivative, multiplies them together and posits the value derived thereof to be the answer.

 

[haruspex]

The book defined alpha (angular acceleration, I take it) merely as $$\frac{dw}{d\theta}$$.

That is a typo in the book. It should by dt in the denominator.

The exact answer has the line in it:
$$\alpha=\omega \frac{d\omega}{d\theta}$$ .

That is correct, but you omitted the / when you wrote alpha=w(dw)(dtheta).