I to start with this Rotations problem

[Sarah2325]

Homework Statement

a tilted plane with no friction with an angle "u" rotates around the Zo axis with constant angular velocity Omega. A kirby (mass = m) starts at the origin resting, and starts going down accelerating. Aproximating all terms with (Omega)^2 to zero:
a) write the motion equations for the x,y and z axis
b)Determinate the Normal Force

Relevant Equations

F = F(rot)+2mΩ X V(rot) + mΩ X Ω X r

I have been stuck with this problem since the start of the week, and i don't know how i should start, any help is apreciated

 

[berkeman]

Welcome to PF, Sarah.

What's a Kirby? Is that like a mass sliding with friction? Or maybe without friction? And where is the angle "u" that you mention? I see an angle $\alpha$ but no angle "u" so far...

Also, can you show us some of the work you have done so far to try to start this problem? Thanks.

 

[Sarah2325]

Hi thanks for the answer! i have to apologize since english is not my first language so i forgot some things,
Kirby is just the object in the picture, sorry for not being clear. and the angle is indeed "alpha" but i made a mistake putting "u"
I have tried doing something, but I am completely stuck, since i have problems starting this, I am really sorry if i seem like a beggar with this, but i don't know what else i can do

 

[berkeman]

So the mass is sliding down the frictionless inclined plane, and the inclined plane is rotating Counter Clockwise about the vertical Zo axis? And you need to write the equation for the motion of the mass versus time given the starting orientation of the inclined plane?

 

[Sarah2325]

yes! write the equation of motion for each axis. thanks for answering! I am wondering how to do that since the rotation is really tricky, at least for me

 

[berkeman]

The horizontal component of the vector normal force on the mass points down the inclined plane, so that vector rotates along with the inclined plane. Can you write the equation for the normal force vector components as a function of time? Then the motion of the mass will be influenced by that horizontal component of the normal force, which will tend to accelerate it in a curved path. You should be able to express the acceleration, velocity and position of the mass as a function of time based on the changing horizontal normal force vector component.

Since the acceleration vector is not constant (in direction), you probably can't use the simple equations for acceleration, velocity and position, but you still should be able to calculate those quantities.

 

[berkeman]

BTW, there may be a simpler way to do this using a rotating reference frame that is fixed to the rotating inclined plane. But others will have to help you with that version...

 

[ergospherical]

Relative to the rotating frame $F = Oxyz$,
i.e. $\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}$
$\mathbf{v} = \dfrac{d\mathbf{r}}{dt} \bigg{|}_F$ and $\mathbf{a} = \dfrac{d\mathbf{v}}{dt} \bigg{|}_F$
where $\dfrac{d\boldsymbol{\gamma}}{dt} \bigg{|}_F := \dfrac{d\gamma_x}{dt} \mathbf{i} + \dfrac{d\gamma_y}{dt} \mathbf{j} + \dfrac{d\gamma_z}{dt} \mathbf{k}$
then the vector equation of motion is$$m\left[ \dot{\boldsymbol{\Omega}} \times \mathbf{r} +2\boldsymbol{\Omega} \times \mathbf{v} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{r}) + \mathbf{a}\right] = m\mathbf{g} + \mathbf{N}$$this comprises three equations. Note that $\dot{\boldsymbol{\Omega}} = 0$ in this problem, so you can ignore the first term.

 

[Sarah2325]

sorry if this sounds a little dumb, but how can i get r? to use the equation i mean. Either way thank you!

 

[ergospherical]

You should first try to write the vectors $\boldsymbol{\Omega}$, $\mathbf{g}$ and $\mathbf{N}$ in terms of the rotating basis $\{ \mathbf{i}, \mathbf{j}, \mathbf{k} \}$. For instance, $\boldsymbol{\Omega} = \Omega \cos{\alpha} \mathbf{k} - \Omega \sin{\alpha} \mathbf{i}$. Just leave $\mathbf{r} = x \mathbf{i} + y\mathbf{j} + z\mathbf{k}$ as is, and the same for $\mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k}$ and $\mathbf{a}= a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k}$. Simplify the various cross products and finally write down three equations by equating the coefficients of $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ in turn.

 

[TSny]

Also, note that you are allowed to neglect terms containing $\Omega^2$.

 

[Leo Liu]

I don't really think it is the only method that we find the acceleration under a rotating frame, as we can solve it under cylindrical coordinates.

Let's put the z axis of the cylindrical c. on the vertical surface of the slide, which points upwards, the r axis on the bottom surface of the slide, orthogonal to the z axis, and $\theta$ on the moving ball, in the direction of $\vec\Omega\times\vec r$. The position of a point in the coordinates have the following form $\vec R=f(\theta, r,\, z)$. Please note here that while $\dot\Omega=\theta$, ${\vec\Omega}\neq \dot{\vec\theta}$. The reason for this is that neither angular velocity nor angular displacement is a genuine vector. Physicists define them that way to simplify calculations and to improve our understandings.

Looking at the vertical component of the total force on the object we obtain the following equation:
$$F_z=N\cos\alpha-mg$$
Then we write out the formula of acceleration under polar coordinates. In this case we extend this formula to cylindrical coordinates by adding a z component, as we did in the equation above. Here, we are only concerned with the acceleration in the $r\theta$ plane.
$$\ddot{\vec r}=(\ddot r-r\dot\theta^2)\hat r+(r\ddot\theta +2\dot r\dot \theta)\hat\theta$$
It is obvious that the ball does not have any acceleration in the $\vec\theta$ direction, so we remove the term with $\hat\theta$. We get
$$\ddot{\vec r}=(\ddot r-r\dot\theta^2)\hat r$$
Since the force component in the r direction is $F_r=N\sin\alpha$ , it follows that
$$\frac{N\sin\alpha}{m}={\ddot r-r\Omega^2}$$

Yet we need to find a way to represent N as other known constants/variables.
We begin this process with using the constraints given by the tilted plane.
$$F_z=F_r\tan\alpha\implies N\cos 2\alpha =mg\cos\alpha\implies N=\frac{mg\cos\alpha}{\cos 2\alpha}$$
We've achieved our goal. Now let's sub it back into the equation concerning the acceleration:
$$\frac{(mg\cos\alpha / \cos 2\alpha)\sin\alpha}{m}=\ddot r-r\Omega^2$$
The last step is solving this second order differential equation and is left as an exercise to readers.
After solving, you get r, and thus get the parametrized coordinates wrt. time. Now convert it to a Cartesian coordinates sharing the z axis. Then use a 3D rotation matrix to convert the position into the coordinates your prof. gave you.