# Moving in an non-inertial frame

• sergiokapone
In summary: 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.
sergiokapone

## Homework Statement

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.

## Homework 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]##

## 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

Hello, Sergiokapone.

sergiokapone said:
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]##

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

Projected on the axis:
##
\frac{{d{v_y}}}{{dt}} = 0##

How did you get the right side to be 0?

Last edited:
TSny said:
How did you get the right side to be 0?

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

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.

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

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.

TSny said:
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.

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}##.

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##

1 person
TSny said:
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##

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

sergiokapone said:
Oh, right. But this mean that the rotation of carousel is decelereted.

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.

## 1. What is a non-inertial frame?

A non-inertial frame is a reference frame that is accelerating or rotating. In this frame, Newton's laws of motion do not hold true and additional forces, such as centrifugal or Coriolis forces, need to be considered.

## 2. How does motion in a non-inertial frame differ from motion in an inertial frame?

In an inertial frame, objects move at a constant velocity unless acted upon by an external force. In a non-inertial frame, objects may appear to accelerate or change direction without any external forces acting on them due to the frame's acceleration or rotation.

## 3. What is the Coriolis effect in a non-inertial frame?

The Coriolis effect is a deflection of moving objects in a non-inertial frame due to the frame's rotation. This effect is responsible for the rotation of hurricanes and the movement of objects on a spinning carousel.

## 4. How does the centrifugal force affect motion in a non-inertial frame?

The centrifugal force is a perceived force that appears to push an object away from the center of rotation in a non-inertial frame. This force is a result of the frame's acceleration and can cause objects to move in a curved path.

## 5. Are there any real-life applications for understanding motion in a non-inertial frame?

Yes, understanding motion in a non-inertial frame is crucial for various fields such as aerospace engineering, oceanography, and geophysics. It helps in predicting the behavior of objects in rotating or accelerating systems, such as spacecraft, ocean currents, and Earth's rotation.

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