# Moving in an non-inertial frame

1. Dec 14, 2013

### sergiokapone

1. The problem statement, all variables and given/known data

In opposite points carousel diameter D = 20 m, rotating with constant angular acceleration, located at the point C shooter and target M. Shooter aiming at a target without introducing amendments to the rotation of the carousel. What should be the angular acceleration of the carousel with that under these conditions, the bullet hit the target at the time the shot was carousel angular velocity $\omega_0$ = 1 rad / min, and the velocity $v_0$ = 200 m / s. Shoter and shooting conditions are assumed ideal. Neglect the influence of centrifugal force.

2. Relevant equations
The equation of motion in non-inertial reference frame
$\frac{{d\vec v}}{{dt}} = - \left[ {\vec \beta \times \vec r} \right] + 2\left[ {\vec v \times \vec \omega } \right]$

3. The attempt at a solution

Projected on the axis:
$\begin{array}{l} \frac{{d{v_y}}}{{dt}} = 0\\ \frac{{d{v_x}}}{{dt}} = 2{v_y}\omega + \beta y \end{array}$

and

$\omega = {\omega _0} + \beta t$
$y = {v_0}t - \frac{D}{2}$ - from the first equation of motion

I finally got
$\frac{{d{v_x}}}{{dt}} = 2{v_0}\omega + 3{v_0}\beta t - \frac{{\beta D}}{2}$

integrating this equation, I get

$x=v_0 \omega_0 t^2+1/2v_0\beta t^{3}-1/4\beta D{t}^{2}$

From the $y = {v_0}t - \frac{D}{2}$ I find time, when the bullet hit the target ($y=D/2$)
$\tau=D/v_0$
when the bullet hit the target $x=0$

solving $0=v_0 \omega_0 t^2+1/2v_0\beta t^{3}-1/4\beta D{t}^{2}$
I find the angular acceleration. $\beta=-\frac{4v_0\omega_0}{D}$
But it is negative!
Help me find the error!

Picture

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 14, 2013

### TSny

Hello, Sergiokapone.

Is the sign of the first term on the right correct? [EDIT: Sorry, I think your sign is correct!]

How did you get the right side to be 0?

Last edited: Dec 14, 2013
3. Dec 14, 2013

### sergiokapone

There is no forces acts in y-direction (I neglect the centrifugal force due to problem condition)

4. Dec 14, 2013

### TSny

In the rotating non-inertial frame the pseudo-forces $\vec{r} \times \vec{\beta}$ and $2\vec{v} \times \vec{\omega}$ will have y-components.

5. Dec 14, 2013

### TSny

Are you making an approximation? Since $x$ and $v_x$ of the bullet in the rotating frame will be very small during the flight, then you can neglect the y-component of the acceleration of the bullet in the rotating frame.

If so, then I think your analysis is correct. The negative value of the angular acceleration $\beta$ is what you should expect. Try to picture the rotational motion of the carousel.

6. Dec 14, 2013

### sergiokapone

But, if value of the angular acceleration $\beta$ is negavive, then I can simply use expression for angular velocity $\omega=\omega_0-\beta t$ instead of with "+" sign expression $\omega=\omega_0+\beta t$, and I should expect the same but positive value of $\beta$, but it is not. With $\omega=\omega_0-\beta t$ I give $\frac{12v_0\omega_0}{5D}$.

7. Dec 14, 2013

### TSny

You should get the same magnitude for $\beta$.

Don't forget to change the sign of $\beta$ in the last term of $\frac{{d{v_x}}}{{dt}} = 2{v_y}\omega + \beta y$

8. Dec 15, 2013

### sergiokapone

Oh, right. But this mean that the rotation of carousel is decselereted.

9. Dec 15, 2013

### TSny

That's right, in the inertial frame the carousel slows down and switches direction so that the target can get back to where the bullet will strike.

There are other solutions where the constant acceleration is positive. But these solutions are not realistic and you would definitely not be able to make the assumption $a_y \approx 0$ in the rotating frame for these other solutions.