Particle moving in a rotating disc

[Like Tony Stark]

Homework Statement

The particle $A$ moves along the groove while the disc turns about $O$ with an angular velocity $\omega$. Find the $x$ and $y$ components of the acceleration of $A$ with respect to Earth when $\omega =3 rad/s ; \dot{\omega}=-10 rad/s^2 ; x=7.5 cm ; v_x =10 cm/s$ and $a_x =15 cm/s^2$.

Relevant Equations

$\vec a=\vec a_B + \vec{\dot \omega} \times \vec r + \vec \omega \times (\vec \omega \times \vec r) + 2. (\vec \omega \times \vec v_{rel}) + \vec a_{rel}$

Well, I tried plugging the data in the formula. I know that $\vec a_b = 0; \vec \omega=3 rad/s ; \vec r$ can be calculated using trigonometry. Then I also know that $v_{relx}= 10 cm/s$, $a_{relx}=15 cm/s^2$, $\vec {\dot{\omega}}=-10 rad/s^2$.

But how do I get $v_{rely}$ and $a_{rely}$? And what's the difference between $\vec a_{rel}$ and $\vec a$?

 

[TSny]

Well, I tried plugging the data in the formula. I know that $\vec a_b = 0; \vec \omega=3 rad/s ; \vec r$ can be calculated using trigonometry. Then I also know that $v_{relx}= 10 cm/s$, $a_{relx}=15 cm/s^2$, $\vec {\dot{\omega}}=-10 rad/s^2$.

You need to specify directions for the vectors $\vec \omega$ and $\dot {\vec \omega}$.

But how do I get $v_{rely}$ and $a_{rely}$?

Think about the time dependence of $y_{rel}$.

And what's the difference between $\vec a_{rel}$ and $\vec a$?

What are the two frames of reference that you are working with in this problem? $\vec a_{rel}$ is the acceleration of $A$ relative to one of these frames and $\vec a$ is the acceleration of particle $A$ relative to the other frame. You need to decide which acceleration goes with which frame. The source from which you got the "Homework Equation" probably describes the notation.

 

[Like Tony Stark]

You need to specify directions for the vectors $\vec \omega$ and $\dot {\vec \omega}$.

Think about the time dependence of $y_{rel}$.

What are the two frames of reference that you are working with in this problem? $\vec a_{rel}$ is the acceleration of $A$ relative to one of these frames and $\vec a$ is the acceleration of particle $A$ relative to the other frame. You need to decide which acceleration goes with which frame. The source from which you got the "Homework Equation" probably describes the notation.

What do you mean with "Think about the time dependence of $y_{rel}$"? It seems that the velocity doesn't have $y$ component if you consider that the ball just moves from side to side. But it confuses me when I think about the rotation of the disc. Does it have $y$ component in that case?

 

[TSny]

What do you mean with "Think about the time dependence of $y_{rel}$"? It seems that the velocity doesn't have $y$ component if you consider that the ball just moves from side to side. But it confuses me when I think about the rotation of the disc. Does it have $y$ component in that case?

Yes, it can be confusing. It's not clear whether the axes drawn in the picture are fixed relative to the disk or fixed relative to the inertial frame. So, to make things clear, imagine x and y axes are painted on the disk in the orientation shown in the figure. Call these axes the x_rel-axis and the y_rel-axis. These axes rotate with the disk such that the x_rel-axis always remains parallel to the groove.

The position of particle $A$ relative to the disk can be specified by the coordinates $x_{rel}$ and $y_{rel}$ of the particle relative to these axes. $v_{rel y}$ is the time derivative of $y_{rel}$. So, if you know the time dependence of $y_{rel}$, you can deduce the value of $v_{rel y}$. That's why I suggested thinking about the time dependence of $y_{rel}$. It does turn out to be pretty trivial.

 

[Like Tony Stark]

Yes, it can be confusing. It's not clear whether the axes drawn in the picture are fixed relative to the disk or fixed relative to the inertial frame. So, to make things clear, imagine x and y axes are painted on the disk in the orientation shown in the figure. Call these axes the x_rel-axis and the y_rel-axis. These axes rotate with the disk such that the x_rel-axis always remains parallel to the groove.

The position of particle $A$ relative to the disk can be specified by the coordinates $x_{rel}$ and $y_{rel}$ of the particle relative to these axes. $v_{rel y}$ is the time derivative of $y_{rel}$. So, if you know the time dependence of $y_{rel}$, you can deduce the value of $v_{rel y}$. That's why I suggested thinking about the time dependence of $y_{rel}$. It does turn out to be pretty trivial.

Oh, I see... but the data that was given to me is with respect to what? With respect to the disc, as if it was stationary? Or respect to the Earth, as if I was seeing the disc in a stationary place?

 

[TSny]

Oh, I see... but the data that was given to me is with respect to what? With respect to the disc, as if it was stationary? Or respect to the Earth, as if I was seeing the disc in a stationary place?

I think the velocity and acceleration components, $v_x$ and $a_x$, given in the problem are relative to the disk, not the earth. They could have been clearer about that.

They ask you to find the acceleration relative to the earth.