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Physics related Edits

In this section, I'll be keeping track of all major physics related edits I've made and will also try to outline the reasons for the edits.

In Uniform Circular Motion

Original:

The above picture shows a point of mass that is moving with a constant angular speed around a center. When the change in angle is ${\displaystyle \Delta \theta }$ , the change in displacement is ${\displaystyle \Delta }$s. Using the relationship of trigonometric functions, we find that,
${\displaystyle \Delta s={\frac {r\sin \Delta \theta }{\cos(\Delta \theta /2)}}}$

This equation is only valid when ${\displaystyle \Delta \theta }$ does not equal ${\displaystyle (2n+1)\pi }$ where n is integer.
Similarly, the magnitude of tangential speed is always the same. Let ${\displaystyle \Delta v}$ be the change in velocity, v be the initial velocity or instantaneous velocity, and ${\displaystyle \Delta t}$ be the change in time,
${\displaystyle \Delta v={\frac {v\sin \Delta \theta }{\cos(\Delta \theta /2)}}}$
${\displaystyle {\frac {\Delta s}{\Delta t}}={\frac {r\sin \Delta \theta }{\Delta t\cos(\Delta \theta /2)}}}$

When ${\displaystyle \Delta t\to 0}$, ${\displaystyle \Delta \theta \to 0}$,

${\displaystyle \lim _{{\Delta t}\to 0}{\frac {\Delta s}{\Delta t}}=\lim _{\Delta t\to 0}{\frac {r\sin \Delta \theta }{\Delta t\cos(\Delta \theta /2)}}}$

${\displaystyle v=\lim _{\Delta t\to 0}{\frac {r\Delta \theta }{\Delta t}}}$
${\displaystyle v=r{\frac {d\theta }{dt}}=r\omega }$ (${\displaystyle \omega }$ is angular speed)
${\displaystyle {\frac {r}{\Delta s}}={\frac {\cos(\Delta \theta /2)}{\sin \Delta \theta }}={\frac {v}{\Delta v}}}$
${\displaystyle \Delta v={\frac {v\Delta s}{r}}}$
${\displaystyle {\frac {\Delta v}{\Delta t}}={\frac {v\Delta s}{r\Delta t}}}$
${\displaystyle \lim _{\Delta t\to 0}{\frac {\Delta v}{\Delta t}}=\lim _{\Delta t\to 0}{\frac {v\Delta s}{r\Delta t}}}$
${\displaystyle a={\frac {v^{2}}{r}}}$

Because ${\displaystyle v=r\omega \!}$
We can substitute into ${\displaystyle a={\frac {v^{2}}{r}}}$
to get: ${\displaystyle a={\frac {(r\omega )^{2}}{r}}=r\omega ^{2}}$

Changed:

Magnitude

Assume ${\displaystyle \omega \!}$ is the angle in radians the body covers in unit time - in other words, the angular velocity of the body. Note that the rotational motion being uniform here we may choose any arbitrary unit for time with no change to the value of ${\displaystyle \omega \!}$ - so, without loss of generality, we assume that our unit of time is infinitesimally small.

The velocity vector, being always tangential to the circle, also turns by an angle ${\displaystyle \omega \!}$ in unit time.

The magnitude of instantaneous acceleration, by definition ,is the magnitude of vector change in velocity in an infinitesimally small period of time. The vector representing this is represented in blue in the above diagram. Since our unit of time is infinitesimally small, for all finite velocities, so will be ${\displaystyle \omega \!}$. Hence, the magnitude of the acceleration vector is ${\displaystyle v\omega \!}$.

Thus, we can conclude that the instantaneous acceleration experienced by a body in uniform circular motion is ${\displaystyle v\omega \!}$.

We may substitute ${\displaystyle v=r\omega \!}$ in ${\displaystyle a=v\omega \!}$ to get:
${\displaystyle a={\frac {v^{2}}{r}}}$

Direction

Since the magnitude of velocity never changes, the direction of the acceleration vector can only be perpendicular to the velocity vector. It can be seen from the diagram that this acceleration vector must point to the center of the circle.

Reason for Edit:

The original derivation was unnecessarily complex and leaves the reader with no feel for the underlying physics. I believe this derivation is direct and elegant - and the same time lets the reader see directly why the magnitude is v*v/r.

In Buoyancy

Original:

Pressure increases with depth below the surface of a liquid. Any object with a non-zero vertical depth will see different pressures on its top and bottom, with the pressure on the bottom being higher. This difference in pressure causes the upward buoyancy force.

The hydrostatic pressure at a depth h in a fluid is given by

${\displaystyle P=\rho hg\,}$

where

${\displaystyle \rho \,}$ is the density of the fluid,

${\displaystyle h\,}$ is the depth (negative height), and

${\displaystyle g\,}$ is the standard gravity (${\displaystyle \scriptstyle \approx \,}$ -9.8 N/kg on Earth)

The force due to pressure is simply the pressure times the area. Using a cube as an example, the pressure on the top surface (for example) is thus

${\displaystyle F_{\mathrm {top} }=d^{2}\rho h_{\mathrm {top} }g\,}$

where ${\displaystyle d}$ is the length of the cube's edges. The buoyant force is then the difference between the forces at the top and bottom

${\displaystyle F_{\mathrm {buoyancy} }=d^{2}\rho h_{\mathrm {top} }g-d^{2}\rho h_{\mathrm {bottom} }g\,}$

which reduces to

${\displaystyle F_{\mathrm {buoyancy} }=d^{2}\rho g({h_{\mathrm {top} }}-{h_{\mathrm {bottom} }})\,}$

in the case of a cube, the difference in ${\displaystyle h\,}$ between the top and bottom is ${\displaystyle -d\,}$, so

${\displaystyle F_{\mathrm {buoyancy} }=-d^{3}\rho g\,}$

or

${\displaystyle F_{\mathrm {buoyancy} }=-\rho Vg\,}$

where V is the volume of the cube, ${\displaystyle d^{3}\,}$

The negative magnitude implies that it is in the opposite direction to gravity. It can be demonstrated mathematically that this formula holds true for any submerged shape, not just a cube.

Changed:

The magnitude of buoyant force may be appreciated from the following argument. Consider any volume of liquid of arbitrary shape and volume ${\displaystyle V\,}$. The body of liquid being in equilibrium, the net force the surrounding body of liquid exerts on it must be equal to the weight of that volume of liquid and directed opposite to gravitational force. That is, of magnitude:

${\displaystyle \rho Vg\,}$ , where ${\displaystyle \rho \,}$ is the density of the liquid, ${\displaystyle V\,}$ is the volume of the body of liquid , and ${\displaystyle g\,}$ the standard gravity (${\displaystyle \scriptstyle \approx \,}$ -9.8 N/kg on Earth)

Now, if we replace this volume of liquid by a solid body of the exact same shape, the force the surrounding body of liquid exerts on it must be exactly the same as above. In other words the "buoyant force" on a submerged body is directed in the opposite direction to gravity and is equal in magnitude to : ${\displaystyle \rho Vg\,}$ ( note that here ${\displaystyle V\,}$ is the volume of fluid displaced by the body )

Reason for Edit:

The original derivation was unnecessarily lengthy and leaves the reader with no real understanding of the underlying physics. I believe this derivation is direct and elegant - and the same time lets the reader feel the physics behind it.

Special Relativity and Time Dilation.. a thought

Pertinent Article:Time Dilation

Just a thought.. it would be completely valid to presume as foundational postulate to special relativity that all objects age/move the same in space-time (dt) ^2 +(ds)^2 is exactly the same for all objects observed in a particular inertial frame.. where dt is how much i percieve the body to have "aged".. while ds is how much i percieve the body to have "moved" spatially in that interval. Distance obviously being measured in 'c' units. Needless to say, time dilation results immediately follow from that assumption. A more elegant way of understanding the result i feel than the usual textbook derivation. Lets the reader appreciate that space and time are not two separate entities but things we percieve as seperate - a limitation imposed by human senses. While the concept space-time is what truly makes sense.

time dilation( and other) results follow intuitively.. ( clock (1) at rest vs clock(2) moving at velocity v:

(dt1)^2 = (dt2)^2 +((v*dt1)/c )^2

dt2=dt1*sqrt(1-(v/c)^2)

Dilip rajeev (talk) 22:30, 2 July 2008 (UTC)

Human Rights Torch Relay

If you have some time please provide us with an input at this RFC on 2008 Summer Olympics torch relay article and this Merger Contest. Thank You! --HappyInGeneral (talk) 00:00, 3 September 2009 (UTC)

RfC/User on PCPP

Hello. Please be aware that I have opened an RfC about the conduct of PCPP (talk · contribs).--Asdfg12345 01:12, 2 March 2010 (UTC)

ANI on your behavior

Hi, I've opened a case on your recent editing behavior here.--PCPP (talk) 13:23, 6 March 2010 (UTC)

==File deletion

File:Persecutionofzhangzhong.jpg listed for deletion

A file that you uploaded or altered, File:Persecutionofzhangzhong.jpg, has been listed at Misplaced Pages:Files for deletion. Please see the discussion to see why this is (you may have to search for the title of the image to find its entry), if you are interested in it not being deleted. Thank you. Ouyuecheng (talk) 17:54, 8 March 2010 (UTC)

WP:AE

Kindly note that I have opened an Arbitration Enforcement case in your name today. Ohconfucius 05:31, 10 March 2010 (UTC)