# Falling body in a non-inertial reference frame

1. Dec 18, 2007

### karnten07

1. The problem statement, all variables and given/known data
Consider a body falling from a tower in the northern hemisphere at approx 40 degrees latitude. The body is seen to have a displacement to the east. Explain the origin of this displacement qualitatively from the point of view of a non-inertial observer. What local effect is responsible for this phenomenon.

[Hint: recall the righthand rule]

2. Relevant equations

3. The attempt at a solution

The rotation of the earth causes acceleration of the local frame. An inertial force throws the body outwards away from the earth's axis of rotation.

I don't understand how to apply the right hand rule here and how this shows me that the body has a displacement to the east.

2. Dec 19, 2007

### Shooting Star

What are the inertial forces acting on the body in a rotating frame? Write down the names and the mathematical expressions. You'll undersatnd.

3. Dec 20, 2007

### Syd12107

Perhaps this might help?

Hi,

Perhaps the following experiment might help.

Think about being on a merry-go-round. If you are close to the center and leaped straight up into the air, your horizontal speed would not be very great - you would land on the merry-go-round almost at the point from where you jumped. However, if you are near the edge of the merry-go-round and jumped upwards, your horizontal speed would be much greater since the edge of the merry-go-round is moving faster than a point near the center (via $$v_T=r\omega$$ where $$v_T$$ is the tangential velocity, $$r$$ is the radius, and $$\omega$$ is the angular velocity). When you come down from your leap, you most likely will not land on the merry-go-round because your faster horizontal velocity will have carried you off the edge. In other words, the farther you are from the axis of rotation, the faster the tangential velocity will be (at the same angular velocity of the merry-go-round, of course).

Now transfer this experiment to a tower on the Earth. At the Earth's surface, the tangential velocity has some numerical value given by $$v_T(R)=R\omega$$ where $$R$$ is now the length of the line perpendicular to the axis of rotation at your latitude value, say $$\theta$$. Then $$R=r_\text{Earth}\cos\theta$$ where $$r_\text{Earth}$$ is the radius of the Earth. If the tower has a height of $$h$$, then the top of the tower is farther from the axis of rotation than its base, now given by $$R_h=(r_\text{Earth}+h)\cos\theta$$. Note that $$R_h>R$$. Therefore the tangential velocity at the top of the tower is greater than at its base. To be specific, the equations give the tangential velocity at the top $$v_T(R_h)=R_h\omega>R\omega=v_T(R)$$ greater than the tangential velocity at the base. So an object dropped from the top of a tower is traveling faster tangentially than the base of the tower. Hence, as it falls, its horizontal velocity (neglecting air resistance) must stay the same as it was at the tower top, and this horizontal velocity is greater than the tower base's horizontal velocity. Thus the object "outruns" the base --- that is, it falls towards the East!

A proper mathematical description of this effect will involve a term looking like $$-2\vec{\omega}\times\vec{v}$$ where $$\times$$ is not the normal multiplication between two real numbers but rather is the cross product between two vectors: the $$\vec{\omega}$$ vector is the vector pointing in the direction of the axis of rotation (towards Polaris, in other words) and $$\vec{v}$$ is the vector pointing in the direction of the velocity of the object, in this case in the vertical downward direction. The cross product of two vectors is perpendicular to both, and by the right-hand rule, would point in the Eastward direction in both the Northern and Southern Hemispheres. So, an object dropped from a tower falls East of the vertically straight down location. By the way, this is the same reason why if you shoot a rifle vertically straight up, the bullet will land to the West of your location -- believe it or not, this little experiment was performed at the Picatinny Arsenal in NJ in the 1930s, I believe. They shot a rifle vertically in the center of a small lake on a calm day to see where the bullet splashed down --- I believe they never could locate its landing position.

If you want help figuring out how the $$-2\vec{\omega}\times\vec{v}$$ term (called Coriolis) arises, just ask.

Good luck,
Syd

P.S. I'm new at this...hope my answer wasn't too detailed or longwinded.

4. Dec 20, 2007

### Shooting Star

Hi Syd12107,

We appreciate your effort at help and the detailed answer that you have given. But we also want the OPs themselves to make an effort and answer as much as they can, so that that they will understand the logic of it all. In case they can't, then definitely you and I and the others will all pitch in to help.

5. Dec 20, 2007

### uber

The right hand rule is used to determine the direction of an angular momentum vector (in this instance). What would that be?

6. Dec 22, 2007

### Syd12107

Hi ... (RE: Right-hand Rule),

I hope I have interpreted your question correctly - i.e., as a question about what the Right-hand Rule is...

Say you wish to take the cross product of two vectors in 3-space, $$\vec{a}\times\vec{b}=\vec{c}$$, then the resultant vector, $$\vec{c}$$, is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$. In a right-hand coordinate system, the direction of $$\vec{c}$$ will be given by the Right-hand Rule.

Right-hand Rule: Curl your fingers from $$\vec{a}$$ towards $$\vec{b}$$, that is, if the center of your right-hand palm is at the origin of vector $$\vec{a}$$, then curl your fingers in the direction from $$\vec{a}$$ towards $$\vec{b}$$. Stick your thumb out as if attempting to hitch hike, and the direction your thumb is pointing is the direction of $$\vec{c}=\vec{a}\times\vec{b}$$.

Notice that $$\vec{a}\times\vec{b}\ne\vec{b}\times\vec{a}$$, that is, the cross product is not commutative.

Lastly, a right-hand coordinate system for 3-space has $$\boldsymbol{i}\times\boldsymbol{j}=\boldsymbolo{k}$$...in other words, the x, y, and z axes are oriented accoring to the Right-hand Rule. (By convention, most of physics employs a right-hand coordinate system.)

Syd

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Last edited: Dec 22, 2007