How can I find the velocity of a point on a disk rotating in a disc?

[burian]

Homework Statement

see attached

Relevant Equations

$ T = I\alpha$

From a freebody analysis I got,
$$ \vec{r} \times \vec{F} = |r| |F| \sin( 90 - \theta) = (R-r) mg \cos \theta$$

and, this is equal to $$ I \alpha_1$$ where the alpha_1 is the angular acceleration of center of mass of small circle around big one,

$$ I \alpha = (R-r) mg \cos \theta$$

Now, tossing all constants as kappa

$$ \alpha = \kappa \cos \theta$$
and integrating

$$ \frac{ \omega^2}{2} = \kappa \sin \theta +C$$

$$ \omega = \sqrt{ 2 \kappa \sin \theta +C}$$

$$ \frac{ d \theta} { \sqrt{ 2 \kappa \sin \theta +C} } = dt$$

$$\int_{0}^{ 2 \pi} \frac{ d \theta} { \sqrt{ 2 \kappa \sin \theta +C} } = \int_{0}^{T} dt$$

As in one whole revolution one time period is covered, however this integral seemsimpossible. SoI'm not sure how to bypass elliptical integral or if it was my physics, idk where I went wrong

I suppose we could solve that equation for 'C', then get an equation for omega.

But, suppose I have the angular velocity equation, then what would I do with it to find velocity of the point in question?

By the way is angular momentum around centre of small circle, conserved? (the one angular velocity of small circle rotating around it's own axis)

 

[haruspex]

Homework Statement:: see attached
Relevant Equations:: $ T = I\alpha$

View attachment 267751

From a freebody analysis I got,
$$ \vec{r} \times \vec{F} = |r| |F| \sin( 90 - \theta) = (R-r) mg \cos \theta$$

and, this is equal to $$ I \alpha_1$$ where the alpha_1 is the angular acceleration of center of mass of small circle around big one,

$$ I \alpha = (R-r) mg \cos \theta$$

Now, tossing all constants as kappa

$$ \alpha = \kappa \cos \theta$$
and integrating

$$ \frac{ \omega^2}{2} = \kappa \sin \theta +C$$

$$ \omega = \sqrt{ 2 \kappa \sin \theta +C}$$

$$ \frac{ d \theta} { \sqrt{ 2 \kappa \sin \theta +C} } = dt$$

$$\int_{0}^{ 2 \pi} \frac{ d \theta} { \sqrt{ 2 \kappa \sin \theta +C} } = \int_{0}^{T} dt$$

As in one whole revolution one time period is covered, however this integral seemsimpossible. SoI'm not sure how to bypass elliptical integral or if it was my physics, idk where I went wrong

I suppose we could solve that equation for 'C', then get an equation for omega.

But, suppose I have the angular velocity equation, then what would I do with it to find velocity of the point in question?

By the way is angular momentum around centre of small circle, conserved? (the one angular velocity of small circle rotating around it's own axis)

Your whole approach is unnecessarily complicated. You do not care about forces or torques. It is a kinematic problem.

At any instant, what is the speed of the disc's centre?

 

[BvU]

if it was my physics

there is no physics involved. It's math.
Simpler problem: disk rolls without slipping on a flat surface with velocity $v$

[edit] ha, hallo haru !

 

[etotheipi]

When I saw torques in your relevant equations, I presumed the configuration was vertical and under the action of gravity. It turns out that for small angular displacements, the disk does undergo simple harmonic motion. Anyway, that is not your problem, but you might find it interesting once you finish this one.

Digression aside, for this problem you only need to know one theorem about rigid bodies (you can probably work it out intuitively, too, but it might help to have the formalism also). That fact is that if an arbitrary point P fixed in the rigid body has a velocity $\vec{v}_P$ in some inertial frame $\mathcal{F}$, then another point Q on the rigid body whose position with respect to P is $\vec{r}'$ will have a velocity in $\mathcal{F}$ of$$\vec{v}_Q =\vec{v}_P + \vec{\omega} \times \vec{r}'$$if $\vec{\omega}$ is the angular velocity of the rigid body (it's because if you write $\vec{r}_Q = \vec{r}_P + \vec{r}'$ and differentiate those in $\mathcal{F}$ i.e. $\left( \frac{d\vec{r}_Q}{dt} \right)_{\mathcal{F}} = \left( \frac{d\vec{r}_P}{dt} \right)_{\mathcal{F}} + \left( \frac{d\vec{r}'}{dt} \right)_{\mathcal{F}}$, you can use that $\left( \frac{d\vec{r}'}{dt} \right)_{\mathcal{F}} = \vec{\omega} \times \vec{r}'$ because $\vec{r}'$ is constant in the body fixed frame)

 

[burian]

Your whole approach is unnecessarily complicated. You do not care about forces or torques. It is a kinematic problem.

At any instant, what is the speed of the disc's centre?

Mhm but is my approach conceptually correct or not?

 

[burian]

When I saw torques in your relevant equations, I presumed the configuration was vertical and under the action of gravity. It turns out that for small angular displacements, the disk does undergo simple harmonic motion. Anyway, that is not your problem, but you might find it interesting once you finish this one.

Digression aside, for this problem you only need to know one theorem about rigid bodies (you can probably work it out intuitively, too, but it might help to have the formalism also). That fact is that if an arbitrary point P fixed in the rigid body has a velocity $\vec{v}_P$ in some inertial frame $\mathcal{F}$, then another point Q on the rigid body whose position with respect to P is $\vec{r}'$ will have a velocity in $\mathcal{F}$ of$$\vec{v}_Q =\vec{v}_P + \vec{\omega} \times \vec{r}'$$if $\vec{\omega}$ is the angular velocity of the rigid body (it's because if you write $\vec{r}_Q = \vec{r}_P + \vec{r}'$ and differentiate those in $\mathcal{F}$ i.e. $\left( \frac{d\vec{r}_Q}{dt} \right)_{\mathcal{F}} = \left( \frac{d\vec{r}_P}{dt} \right)_{\mathcal{F}} + \left( \frac{d\vec{r}'}{dt} \right)_{\mathcal{F}}$, you can use that $\left( \frac{d\vec{r}'}{dt} \right)_{\mathcal{F}} = \vec{\omega} \times \vec{r}'$ because $\vec{r}'$ is constant in the body fixed frame)

now how do I find v_p as a function of time? and how do I find omega?

 

[burian]

now how do I find v_p as a function of time? and how do I find omega?

I don't think it is $$\frac{2\pi}{T}$$ because I think angular velocity may change as a function of time

 

[etotheipi]

I dare say what follows is a bit of an unnecessary complication, because you can just write down $\vec{v}_P$ by thinking about what @BvU told you in #3, but in any case:

I think it is to be assumed that, due to the symmetry of the setup, the angular velocity is constant with time. Let P be the centre of mass of the smaller disk, and let this point be translating about the centre of the larger disk at orbital angular velocity $\vec{\Omega} = \Omega \hat{z}$. Let the angular velocity of the smaller disk be $\vec{\omega} = -\omega \hat{z}$. Due to the rolling condition, you know that$$\Omega \hat{z} \times (R-r) \hat{r} - \omega \hat{z} \times r\hat{r} = 0$$It's much easier to write that in terms of the components of $\hat{\theta} =\hat{z} \times \hat{r}$, so try doing that. Now you have the relationship between the orbital and spin angular velocities. Then, you're after the velocity of the point at $\vec{r}' = -r\hat{r}$, i.e.$$\vec{v} = \Omega \hat{z} \times (R-r) \hat{r} - \omega \hat{z} \times (-r)\hat{r}$$

 

[burian]

mhm , could you recommend places where I can study of orbital and angular velocity? I have not heard of these terms.

and, I wish more explanation on the rolling condition

 

[etotheipi]

What I mean by the rolling condition is that the velocity of the point of the disk which is instantaneously in contact with the larger disk must be zero in order for it to be "rolling without slipping". I feel it would be helpful for you to analyse a simpler example of a disk rolling on a flat surface first.

For info on rolling, look here:
https://physics.info/rolling/

For info on general planar kinematics of rigid bodies, and angular velocities, look here:
https://www.brown.edu/Departments/Engineering/Courses/En4/notes_old/RigidKinematics/rigkin.htm

 

[haruspex]

angular velocity may change as a function of time

You are told the motion is uniform.

 

[haruspex]

Mhm but is my approach conceptually correct or not?

It is not, because you are trying to use forces and torques when you have no information on these.

 

[etotheipi]

You are told the motion is uniform.

Whoops, yes that's a bit of a give-away

 

[burian]

It is not, because you are trying to use forces and torques when you have no information on these.

How is this possible, I can't see it happening in a vertical plane so it's a horizontal plane?

 

[haruspex]

How is this possible, I can't see it happening in a vertical plane so it's a horizontal plane?

Maybe horizontal on a frictionless surface, or perhaps part of a machine - doesn't matter how it is achieved, but somehow it is made to move around in a circle at a constant speed without slipping.

 

[burian]

Ah, I got this one now. As haruspex told, all I had to do was forget how the system was configured and just aspect that it some how magically rotates with omega,

after that I just did,

$$ v= 2 \frac{\pi}{T} (R-r)$$

them I did

$$ v_{top}= 2 v_{com} = 4 \frac{ \pi}{T } (R-r)$$

 

[haruspex]

Ah, I got this one now. As haruspex told, all I had to do was forget how the system was configured and just aspect that it some how magically rotates with omega,

after that I just did,

$$ v= 2 \frac{\pi}{T} (R-r)$$

them I did

$$ v_{top}= 2 v_{com} = 4 \frac{ \pi}{T } (R-r)$$

That's it.