How Do Projectiles Deviate in a Rotating Frame at 45°N Latitude?

In summary, a shell is fired due North at an inclination of 45 degrees with initial speed V at an apparent latitude of 45 degrees. Neglecting the curvature of the Earth, it will hit the ground a distance of 2ωV^3/3g'^2 East of the line of longitude through the point of firing. If fired due South under similar conditions, the deviation will be twice as great and to the West. The position vector of the particle at time t is given by \underline{r} (t) = x(t) \underline{i} + y(t) \underline{j} +z(t) \underline{k}, where x(t) is the distance traveled north, z(t) is the distance
  • #1
maxtor101
24
0

Homework Statement


At apparent latitude 45◦N a shell is fired due North at an inclination of 45◦
to the horizontal with initial speed V . Show that, neglecting the curvature of the Earth, the shell will hit the ground at a point a distance [tex] \frac{2 \omega V^3}{3g' ^2}[/tex]
East of the line of longitude through the point of firing. If the shell is fired due South under similar conditions, show that the deviation will be twice as great and to the West.


Homework Equations



I found the following equation in a mechanics book, it was derived from the rotating axes theorem:
[tex]
\underline{r} (t) = \frac{1}{2}\underline{g}'t^2 - \frac{1}{3} \underline{ \omega} \times \underline{g}' t^3 - \underline{ \omega} \times \underline{V} t^2 + \underline{V}t
[/tex]


Where [tex]\underline{r} (t)[/tex] is the position vector of the particle at time t.
[tex]\underline{g}'[/tex] is the apparent gravity vector
[tex]\underline{ \omega}[/tex] is the angular velocity vector of earth
[tex] \underline{V} [/tex] is the velocity vector it is projected with.

The Attempt at a Solution



Well the position vector must be of the form

[tex] \underline{r} (t) = x(t) \underline{i} + y(t) \underline{j} +z(t) \underline{k}[/tex]

Where i,j,k are the unit vectors with the point of projection at the origin as shown in the diagram.

[tex] x(t) [/tex] is the distance traveled north after t seconds.

Therefore [tex] x(t) = Vt cos45 [/tex]
[tex] x(t) = \frac{V}{\sqrt{2}} t [/tex]

[tex]z(t)[/tex] is the distance above the ground after t seconds.

[tex]z(t) = \frac{V}{\sqrt{2}} t - \frac{1}{2}g't^2 [/tex]

Solving for t when z(t) = 0 we get

[tex] t = \frac{V \sqrt{2}}{g'}[/tex]

This is the time when the particle hits the ground.

Now this is the part that I'm not sure of...

[tex] y(t) [/tex] is the distance Earth rotates underneath the particle as it is in the air.

I'm not sure how to write down [tex] y(t) [/tex]

I know that you find [tex] y(t) [/tex] at [tex] t = \frac{V \sqrt{2}}{g'}[/tex]

I know it may be confusing but any help would be greatly appreciated guys and girls.

Thanks a mill in advance

Max
 
Physics news on Phys.org
  • #2
Sorry, forgot to upload diagram of the axes setup and stuff
 

Attachments

  • diagram.png
    diagram.png
    2.7 KB · Views: 442

Related to How Do Projectiles Deviate in a Rotating Frame at 45°N Latitude?

1. How does the rotation of a frame affect the motion of projectiles?

The rotation of a frame can affect the motion of projectiles in a few ways. First, it can alter the direction of the projectile's initial velocity, causing it to move in a curved path rather than a straight line. Second, the Coriolis effect can cause the projectile to experience a deflection to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. Finally, the centrifugal force can also impact the trajectory of the projectile, causing it to follow a curved path.

2. What is the Coriolis effect and how does it impact projectiles?

The Coriolis effect is a phenomenon that occurs due to the rotation of the Earth. As a projectile moves in a rotating frame, it experiences a deflection to the right in the Northern Hemisphere and to the left in the Southern Hemisphere. This effect is caused by the difference in linear velocities between different points on the Earth's surface due to its rotation.

3. How does the centrifugal force affect projectiles in a rotating frame?

The centrifugal force is a fictitious force that appears to act on objects moving in a curved path in a rotating frame. This force can impact the trajectory of projectiles by causing them to follow a curved path instead of a straight line. The magnitude and direction of the centrifugal force depend on the velocity and position of the projectile within the rotating frame.

4. Can the laws of motion be applied to projectiles in a rotating frame?

Yes, the laws of motion can still be applied to projectiles in a rotating frame. However, the equations for projectile motion will need to be modified to account for the effects of rotation, such as the Coriolis force and the centrifugal force. By incorporating these additional factors, the laws of motion can accurately describe the motion of projectiles in a rotating frame.

5. How does the conservation of angular momentum apply to projectiles in a rotating frame?

The conservation of angular momentum states that the total angular momentum of a system remains constant unless acted upon by an external torque. In the case of projectiles in a rotating frame, the angular momentum of the projectile will remain constant as long as there is no external torque acting on it. This means that the projectile will continue to rotate at a constant rate and follow a curved path due to the effects of rotation.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
926
  • Advanced Physics Homework Help
Replies
7
Views
1K
Replies
26
Views
3K
  • Advanced Physics Homework Help
Replies
12
Views
2K
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
5
Views
2K
Back
Top