Talk:Physics Study Guide/Torque

how about adding (I can't do formulae :) )


 * acceleration toward centre of rotation (centripetal force?) = (scalar velocity)^2/radius = omega^2 x radius
 * Force toward centre of rotation = m x r x omega^2 or (mv^2)/r

--81.157.105.216 19:07, 9 Jan 2005 (UTC)

Torque
I also can't do formulas. There isn't really anything about torque in the torque section! The definition of torque = radius x force should be included at least, right? Dyknowsore

Centripetal Acceleration
In regards to your discussion,

"acceleration toward centre of rotation (centripetal force?) = (scalar velocity)^2/radius = omega^2 x radius"

Keep in mind the formula you listed above states the magnitude of circular acceleration for uniform circular motion. Given that we are dealing with uniform circular motion, we also know that acceleration will always be perpendicular to velocity; thus, we can assume its direction is always towards the center of the circle. If acceleration wasn't perpendicular to the direction of velocity, then the acceleration would be contributing to the velocity in some sort of way. Thus, the magnitude of velocity would not remain constant, thus no longer making it uniform circular motion.

Let: A = Angle R = displacement vector V = velocity vector f = frequency w(omega) = 2piF = velocity/radius (v/r) t = time A = wt i = x-axis unit vector notation j = y-axis unit vector notation

'''1. Displacement vector R is equal to displacement in its x- and y-components. Such that,'''R = R(CosA)i + R(SinA)j -or- 2. Since A=wt, R = R(Cos(wt))i + R(Sin(wt))j [substituting the angle for wt] Note: If you notice, the resultant vector R is equal to the radius at all points. It is basically the radius sweeping out a portion of the circle. In other words, it's only the angle (direction) that changes in uniform circular motion and the magnitude is always equal the radius. For futher proof, you can use the Pythagorean Theorem to show that the magnitude of R equals radius r. '''3. Now deriving R, we produce velocity vector V.''' V=R'=(Rw(-Cos(wt)))i + (Rw(Sin(wt)))j 4. Then we take the second derivative to get acceleration. a(circular)=V'=-(Rw^2(Cos(wt)))i-(Rw^2(Sin(wt))j Then factor the omegas... a(circular)=-w^2(R(Cos(wt)))i+(R(Sin(wt))j 5. Now taking a look at the line above, a(circular)=-w^2(R(Cos(wt)))i+(R(Sin(wt))j the bolded region is equal to vector R, so a(circular)=-w^2(R) 6. And since w=v/r a(circular)=-(v/r)^2(r)= -(v^2/r^2)(R) But we cannot cancel out r, a scalar value, with a vector R.  We know that the magnitude of R is always going to equal r.  So we can take all of the components, square them individually, add them together, and square root them (according to the Pythagorean Theorem).  The value we produce is now in a format we are all used to: Note: The negative sign in front of (v/r)^2 is removed during the squaring process. And thus, we have proved: ~Chris
 * a(circular)|=(v^2/r^2)|R| = (v^2/r^2)r = v^2/r
 * a(circular)| = v^2/r

Centrifugal force
The following statement from the article on 2/5/07 is incorrect:


 * We know that an opposite force should exist for this centripetal force(by Newton's 3rd Law of Motion). This is the centrifugal force, which exists only if we study the body from a non-inertial frame of reference(an accelerating frame of reference,such as in circular motion).

The "equal and opposite force" is usually of the same nature as the centripetal force. For instance, for a uniform planetary orbit, the gravitational force of the planet on the sun is equal and opposite to the force of the sun on the planet. Because of such confusion, I don't think centrifugal force should be discussed in this module -- just mention (correctly) that it is an illusion in the non-inertial frame of reference, and maybe link to a more detailed discussion of the pseudo-force.Zolot 20:55, 5 February 2007 (UTC)