Another "simple" physics question.

I like this thread and I'll tell you why: I see people trying to learn something, and that is very cool.

There's a question above about the difference between speed and velocity. Very good question. They are in fact differnet, and here's how:

Speed is a scalar quantity. That means it has magnitude but no direction. SO if you ask me what my speed is I can look at the speedometer and tell you I am going 60 mph.

Velocity is a vector quantity. To define a velocity you must define both the speed AND the direction. If you ask me what my velocity is, I have to tell you that I am moving 60mph east to fully define my velocity.

Acceleration is a change in velocity. You can accelerate without changing speed. WHAT!?!? Oh man, I'm gonna have to draw a picture......
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So in the diagram above, we see an object with velocity "v" moving north, and an acceleration "a" pushing west. Because the acceleration is perpindicular to the velocity, it changes the velocity of the object without changing its speed. This is the case of a satellite in a circular orbit, or a stink bug on a string. Whatever. The satellite/stink bug, moves in a circle. Its speed is constant, but its velocity is constantly changing, from 60 mph north to 60 mph northwest, to 60 mph west, etc...

 

Spasso, I use the terms speed and velocity interchangealby for everyday conversations I have. Only time I use them differently is in a physics problem, which is the way my professors and books taught me.

Art's diagram shows it well. The ball (or whatever object you put at the end of a rotating string) wants to keep going in a straight line at any point on the circular path (imagine letting the string go and the ball will go in a straight line starting from the point you let go). Since you're holding onto the string, it can't go where it wants and is FORCED to follow the circular path you give it. This force is MassXAcceleration. The ball undergoes acceleration.

 

This formula is wrong. This formula you gave is the one for centripetal acceleration of an object in circular orbit at constant speed about a fixed point. The dimensions (units) are like distance/time^2 (SI : m/s^2)

The formula for instantaneous angular acceleration is in fact, (greek symbol) alpha = a/r, where a is the magnitude of instantaneous linear acceleration and r is the distance to the axis. The units for angular acceleration should be angular measure/time^2 (SI : radians/s^2)

One can also use this relationship : Torque (greek letter tau) = I * alpha, where I is the moment of inertia of the object.

For an object undergoing simple circular motion at constant speed about a fixed point, there is no angular acceleration. It does experience centripetal acceleration, following the formula you gave.

 

Spasso,

The definitions you have quoted are not incorrect, but they are imprecise.

If you read the definition of velocity, they do mention direction by stating it as rate of change of position along a straight line. They could have said "instantaneous rate of change of position along a specific direction."

To their credit, they do differentiate speed from velocity by stating that speed is equal to the magnitude of velocity, irrespective of the direction.

Hope that clears it up.

 

^^I'm sorry, but this is completely and utterly wrong. I applaud the attempt to give a simplistic intro to nonlinear mechanics to someone else, but please get the concepts right.

For starters, there is no such thing as "centrifugal force" in Newtonian mechanics. Newtonian forces only appear in an inertial frame of reference (one that is still or moving with constant velocity). In this sort of frame, there is only a centripetal acceleration and centripetal force.

The centrifugal force (acceleration) is a "fictitious force" or "inertial force" that appears when apparent "forces" are considered in a non-inertial frame of reference. From the point of view of the object that is moving in a circle, there would be an outward throw - this is called "centrifugal force". It is not a Newtonian force, because the object itself is accelerating, and hence does not meet the criterion of inertial frame.

(There is more technical jargon to be said here - there are other famous fictitious forces like the Coriolis "force", and some other technical aspects like D'Alembert's formulation that tries to reformulate Newton's second law of motion to make it fictitious force "friendly". But I digress. Suffice it to say that centrifugal force is not (or at least should not) be taught in a school Physics class, because it can confuse the undiscerning).

For an object moving at a constant speed in a circle, there is no forward component of force or acceleration as you allude to in your blue line. The only acceleration is perfectly centripetal, an inwardly directed vector to the center of rotation. If that force is suddenly removed, the object will move at constant speed at a tangent to the circular path from the point of release. If there were indeed a forward force acting upon it, it would accelerate from that point, but it does not.

When considering a generalised object moving wrt to an origin in two dimensional motion, I always find it easiest to use a complex polar representation. Let the position vector of the object at time t be represented by z(t). Let the distance from axis at time t be r(t) and the angle subtended from the x-axis be theta(t). i represents square root of negative one (these are complex polar coordinates).

So

z = re^(i*theta)

Differentiating twice wrt time to get A, the acceleration of the object we get :

A = d^2(z)/dt^2 = ([a_r] - rw^2)e^(i*theta) + (2[v_r]*w + r@)*i*e^(i*theta)

where [v_r] is the radial velocity, [a_r] is the radial acceleration (respectively, the first and second derivatives of the radial displacement wrt time), "w" is supposed to represent greek omega, the angular velocity, and "@" greek alpha, the angular acceleration.

This is the most general form for the acceleration of a particle. Of the terms on the right hand side, the ones more to the left represent the radial component of the final acceleration and the ones more to the right (after the plus sign) represent the tangential component of the acceleration.

For an object in circular motion at fixed speed, v_r = a_r = 0 and @ = 0, so the expression reduces to :

A = -rw^2*e^(i*theta)

which is in fact a centripetally directed acceleration with magnitude rw^2. Since the tangential velocity v = rw, we can also verify that A = v^2/r, which is the form you're familiar with.

 

Awesome, dude !! My dad had a red 240Z once upon a time ...loved that car. Later he had a Nissan 200SX (a watered down descendant), which I got from him.

 

With all due respect, I am sure you know a good deal more about the matter than I, but I expect the blue line in 2000Yellow360's diagram was meant to show the direction that the object would travel if the inwardly directed force was no longer acting on it. A satellite orbitting the earth experiences the centripetal force of gravity and the forward momentum in the direction indicated by his (presumably?) blue line. I may be wrong here since it has been quite some time since I learned this material in High School.

 

There is no forward/tangential component of force along the blue line if the object is travelling at constant speed in a circle. There is ONLY a centripetal force. There is a tangential component of momentum that would exist if the centripetal force is removed (which is why the object goes "flying off"), but that momentum would not alter unless new forces were introduced. By definition, you need a force to alter momentum.

This is the bit that's completely wrong :

The stuff about summing the vectors to get the resultant grey force vector, that's all also wrong of course. There is only a perfectly inwardly directed force, the centripetal force.

In fact when you take a particle moving at constant speed in a straight line and apply a force at right angles to its path of motion, its speed does not change (magnitude unaltered). Merely the direction of travel changes. This in itself, of course, is enough to constitute a change in velocity, which includes both magnitude and direction. Momentum is also a vector quantity (being the product of mass and velocity), so it changes too.

Circular motion is in fact, the limit of this process, where a force is constantly being applied at right angles to a particle's instantaneous momentum vector. It's that simple, and can easily be verified mathematically.

 

thanks for clearing that up

 

No problem. I edited my post to add more stuff.

 

it would be both accell and then decell at different points on its rotation.... IF the point of refernce is a fixed point

accel and decel are rate of change.

the rate of change in distance from the fixed point on the ground changes as the fixed point on the sweeps through its arc.

 

ft/s/s

 

Maybe now I know why Mr. Payne is laughing about thread.

 

There's no such thing as deceleration in physics. It's either acceleration or negative accleration.

 

Acceleration is a change in speed and/ or direction. There is no negative or positive, it is just plain acceleration. Physical elements of the spinning wheel are in a constant acceleration due to a constant change in direction.

 

common usage of the word....

decel is a dimminishing rate of chagne

 

Okay, but applying the meaning of the word it is still simple acceleration no matter what percieved direction the acceleration is taking place. This forum is one helluva lot of fun and educational. Invigorating to see the interest and involvement....and professionalism in most.....me excluded.

 

Higgens, I used the term loosely. A very common occurance when using the english language, especially when the basic meanings of words are expanded to explain theoretical extrapolations that the common man are unaware of AND COULDN'T CARE LESS ABOUT, LIKE IN THIS THREAD.

The force of the car hitting the tree at the outside of the turn would have been caused by the inertia of trying to navigate the turn at too high of a velocity
Or it could have been caused by a sudden divergence of travel by the vehicle from it's linear travel. Better?

or did I get that wrong too?

 

Car and tree meeting in the turn is a prime example of acceleration due to a constant change in direction. The clinker is the loss of traction of the tires on the the road thus a change in the intitial acceleration when the turn started. Now comes some real acceleration when the car hits the tree at something like 10G -20G. GEE, that hurts.

 

Well, it never happened so I guess it's a moot point but well taken.

 

Art : no response to my gentle corrections ? You agree ? No ?

 

You are correct, but you have complicated it needlessly with all that theta stuff. I skipped over that. I haven't done polar math in eons and never did like it.

The object is moving at a constant velocity in the tangential direction. So, no acceleration in that direction, no force in that direction. I was wrong about that.

The circular motion results from an inward force acting at right angles to the object going at a constant speed (not accelerating tangentially), right?

Thanks for the correction, but I want to make sure I understand you correctly now. I never was very good at this stuff, and what little I learned, I am slowly forgetting. I can still pop a monster wheelie, though. Does that count?

 

You are correct now. Don't worry about the theta stuff, that was just me showing off.