Determine the angular velocity as a function of the angle

[Davidllerenav]

Homework Statement

A solid body begins to rotate around a fixed axis with angular acceleration $\beta=\beta_0\cosφ$, where $\beta_0$ is a constant vector, $φ$, is the angle of rotation of the body from initial position. Determine the angular velocity of this body as a function of the angle $φ$. Represent the graph of this dependence.

Homework Equations

Circular motion.

The Attempt at a Solution

I tried it like this: $\beta_0 \cosφ = \beta = \frac{d\omega}{dt} = \frac{d^{2}φ}{dt^{2}}$. After that I really don't know what to do next. I know that I need to integrate to get the angular velocity, but I don't know how to get to that.

 

[kuruman]

Hint: $\alpha=\frac{d\omega}{dt}=\frac{d\omega}{d\varphi}\frac{d\varphi}{dt}=~?$

 

[Davidllerenav]

Hint: $\alpha=\frac{d\omega}{dt}=\frac{d\omega}{d\varphi}\frac{d\varphi}{dt}=~?$

I don't understand that hint, where does that come from?

 

[SammyS]

I don't understand that hint, where does that come from?

It comes from Calculus.

The Chain Rule.

... along with angular acceleration.

Note that @kuruman is using $\alpha$ for angular acceleration, and $\omega$ for angular velocity.

 

[haruspex]

I don't understand that hint, where does that come from?

It is a standard trick in equations of motion. Working it backwards from the expression on the right, the two dφ terms cancel.
The equation produced by this trick is equivalent to work conservation. Indeed, you can solve the problem by applying that principle.
The trick is most useful when you do not need time in the answer. Note that you are asked to find angular velocity as a function of angle, not of time.

 

[Davidllerenav]

It is a standard trick in equations of motion. Working it backwards from the expression on the right, the two dφ terms cancel.
The equation produced by this trick is equivalent to work conservation. Indeed, you can solve the problem by applying that principle.
The trick is most useful when you do not need time in the answer. Note that you are asked to find angular velocity as a function of angle, not of time.

I see, thanks, so I use tha trick and end up with $\beta_0 \cosφ = \beta = \frac{d\omega}{dt} = \frac{d\omega}{d\varphi}\frac{d\varphi}{dt}=\frac{d^{2}φ}{dt^{2}}$, right?

 

[SammyS]

I see, thanks, so I use tha trick and end up with $\beta_0 \cosφ = \beta = \frac{d\omega}{dt} = \frac{d\omega}{d\varphi}\frac{d\varphi}{dt}=\frac{d^{2}φ}{dt^{2}}$, right?

Not much help in that form.

Better to use: $\ \beta_0 \cos \varphi = \dfrac{d\omega}{d\varphi} \cdot \dfrac{d\varphi}{dt} ~$ .

 

[Davidllerenav]

Not much help in that form.

Better to use: $\ \beta_0 \cos \varphi = \dfrac{d\omega}{d\varphi} \cdot \dfrac{d\varphi}{dt} ~$ .

So I will end up with $\frac{d\omega}{dt}=\beta_0\cos\varphi$, then $d\omega=dt\beta_0\cos\varphi$, right? What do I do next?

 

[kuruman]

So I will end up with $\frac{d\omega}{dt}=\beta_0\cos\varphi$, then $d\omega=dt\beta_0\cos\varphi$, right? What do I do next?

That's not what you end up with. Look at #7 again. What is $\frac{d\varphi}{dt}$ also known as?

 

[Davidllerenav]

That's not what you end up with. Look at #7 again. What is $\frac{d\varphi}{dt}$ also known as?

Oh, sorry it is the angular velocity $\omega$, so I will end up with $\dfrac{d\omega}{d\varphi} \cdot \omega = \beta_0 \cos \varphi$ thus it would be $\omega d\omega = d\varphi \beta_0\cos\varphi$.

 

[kuruman]

Right. Then what?

 

[Davidllerenav]

Right. Then what?

I think that I can integrate it know, right?

 

[kuruman]

Go for it.

 

[Davidllerenav]

Go for it.

What are the limits of the integrals?

 

[Davidllerenav]

Go for it.

I ended up with $\frac{\omega^2}{2} = \beta_0 \sin\varphi$ but it is the result of the indefinite integral.

 

[haruspex]

I ended up with $\frac{\omega^2}{2} = \beta_0 \sin\varphi$ but it is the result of the indefinite integral.

In general, there should be a constant of integration, which you can determine from the initial conditions:

A solid body begins to rotate... φ is the angle of rotation of the body from initial position.

 

[Davidllerenav]

In general, there should be a constant of integration, which you can determine from the initial conditions:

So it would be $\frac{\omega^2}{2}+C = \beta_0 \sin\varphi+C$?

 

[haruspex]

So it would be $\frac{\omega^2}{2}+C = \beta_0 \sin\varphi+C$?

No, you don't put the same constant on both sides, that would be self-defeating. Put it on one side - doesn't matter which - then determine its value.

 

[Davidllerenav]

No, you don't put the same constant on both sides, that would be self-defeating. Put it on one side - doesn't matter which - then determine its value.

Ok, so I'll write it like $\frac{\omega^2}{2} = \beta_0 \sin\varphi+C$. Now, how do I determine its value? Since I don't have the initial conditions, would it be something like $r_0$?

 

[haruspex]

I don't have the initial conditions

You do - see post #16.

 

[Davidllerenav]

You do - see post #16.

Oh, so it would be $C=\varphi$ thus I will have $\frac{\omega^2}{2}= \beta_0 \sin\varphi+\varphi$, right?

 

[haruspex]

Oh, so it would be $C=\varphi$ thus I will have $\frac{\omega^2}{2}= \beta_0 \sin\varphi+\varphi$, right?

No.
According to the text I quoted, what is the velocity when φ=0?

 

[Davidllerenav]

It only says that the body rotates and that $\varphi$ is the angle of rotation of the body in the initial position. I don't know, sorry.

 

[kuruman]

It only says that the body rotates and that $varphi$ is the angle of rotation of the body in the initial position. I don't know, sorry.

It says that it "begins to rotate". What does that suggest?

 

[Davidllerenav]

It says that it "begins to rotate". What does that suggest?

That the velocity is 0?

 

[haruspex]

and that $\phi$ is the angle of rotation of the body in the initial position

No, $\phi$ is the angle of rotation from the initial position. So what is it at the initial position?

 

[haruspex]

That the velocity is 0?

Yes.

 

[Davidllerenav]

Yes.

So the constant would be 0?

 

[haruspex]

So the constant would be 0?

Yes. The given initial condition is that when $\phi=0$ $\omega=0$.

 

[Davidllerenav]

Yes. The given initial condition is that when $\phi=0$ $\omega=0$.

Ok, So it would be $\frac{\omega^2}{2} = \beta_0 \sin\varphi+0$, right? I understand, thanks! How do I draw the graph of the dependence?

 

[haruspex]

Ok, So it would be $\frac{\omega^2}{2} = \beta_0 \sin\varphi+0$, right? I understand, thanks! How do I draw the graph of the dependence?

It just means sketch a graph of ω against φ.

 

[Davidllerenav]

It just means sketch a graph of ω against φ.

But how to I graph that?

 

[haruspex]

But how to I graph that?

How do you sketch any graph?
Pick a few convenient values of x and see what y is.
Same for gradients at the same points.
Consider local extrema and asymptotic behaviour.

 

[Davidllerenav]

How do you sketch any graph?
Pick a few convenient values of x and see what y is.
Same for gradients at the same points.
Consider local extrema and asymptotic behaviour.

But there isn't any x or y. It is polar coordinates, right?

 

[haruspex]

But there isn't any x or y. It is polar coordinates, right?

You are just sketching a graph of one variable against another. The physical meanings are irrelevant.
You could sketch it in polar if you wish, but I doubt that is what is expected.