hi from my understanding force can only change direction of velocity vector when it is perpendicular to it. Meaning once velocity is no longer perpendicular to it it can't change its direction. -If force vector F has a certain angle with velocity vector(not perpendicular) then horizontal component F[h] will be parallel to V vector while vertical F[r] component will be perpendicular to V. Since F[r] is perpendicular to V will it change its direction? I'm asking this because I assumed PERPENDICULAR force changes direction of a velocity vector ONLY in cases when there is no velocity vector's component in same direction as force F or one of component's of force F?! Code: _ _______ /| / / force / thank you
If a mass has a velocity in a certain direction and you apply a force perpendicular to the direction of motion then the mass will stay at constant velocity in that original direction and accelerate perpendicularly to that velocity. The original velocity will not be lost. The mass will follow a curved path of similar shape to the positive side of the graph of equation y=x^2. Is that what you asked? (I couldn't interpret you're diagram in relation to what you said...) Are you relating this question to the forces acting on a mass that is moving in a circular path? If so the original velocity in this case IS lost. Generally if there is a force acting on a mass that is anything but perpendicular to the velocity of that mass then you must convert the force vector into two equivalent component vectors and find the force that acts on the same line as the volocity of the mass. From that point you simply use f=ma.
Then you understand wrong. Whoever said what led to this obviously meant that a perpendicular force can't change the speed, not that you can't change the direction with a force that isn't completely perpendicular. Sometimes you have to consider that you didn't understand what was said in the first place.
I know that. Actually, perpendicular force will slightly change the original direction of velocity vector and only then will it start accelerate perpendicularly(almost) to that original velocity. I thought force only changes the direction of velocity vector when there is no component of velocity vector parallel to force vector? Otherwise force only changes the direction of net velocity vector by changing magnitude of velocity component parallel to the force. If I am wrong, please correct me
You are wrong. A force changes the direction of velocity vector whenever the force has a non-zero component that is not parallel to the velocity vector. Remember that forces add up. Suppose you apply some force normal to the velocity vector. This force bends the velocity vector. Now change that initial normal force by adding a very small force parallel to the velocity vector. That tiny parallel component does not make the consequences of the large normal component vanish. Satellites orbiting the Earth provide a real-world example of this. Even ignoring perturbations from the atmosphere, the Sun, and the Moon, it is impossible to achieve a <i>perfectly</i> circular orbit around the Earth because the Earth has a non-spherical mass distribution. The satellite still orbits the Earth even though the satellite almost always has a component of the gravitational force that is not parallel to the velocity vector.
When you say change the direction are you talking about changing the direction of net velocity vector by adding another component or also actually changing the component perpendicular to velocity vector? So basicly if original horizontal velocity V[h] is perpendicular to force: -force changes a bit the direction of original velocity vector V[h] -Original Velocity vector is no longer perpendicular to force -Force now also adds vertical velocity component to the object -Net velocity gets bigger and constantly changes direction ( object has parabolic path ) -V[h] has still the same magnitude and is no longer horizontal, but now it has horizontal component and this component is perpendicular to force and again force changes the direction of V[h]'s horizontal component... Are the above stages correct?
Any force on any moving mass can be broken up into two components, one parallel to the direction of motion and one perpendicular to the direction of motion. The parallel component will cause a change in the speed. The perpendicular component will cause a change in the direction. -Dale
If you take a charged particle in a magnetic field, at any time the force acting on the particle is perpendicular to the velocity. The direction of the velocity will change but its amplitude will remain the same
This is important to me. Can't someone at least tell me if the steps I wrote two posts above are correct?
Yes, that is the basic idea. When I was talking about breaking up a force into parallel (speed changing) and perpendicular (direction changing) components you must do that separately at each instant in time. The force and velocity have no "memory" it is only the current force that causes a change in the current velocity. As long as you understand that, then you should get the right answer by following the procedure you described. -Dale
I think you'd have to tell us all what an energy vector is first. I've always known energy to be measured in scalar quantities. -AntonK
Energy is not a vector quantity. Vectors have a magnitude and a direction. Energy has no direction. For example, the gasoline in your tank has a certain amount of chemical energy, but what is the direction? More mathematically: W == f . d Where f and d are the force and displacement vectors respectively. The dot product of two vectors is a scalar. -Dale
The Poynting vector is not energy; it is in units of power/area. The directional aspect of the Poynting vector comes from the area (normal vector to a surface), not the power. -Dale
>> The directional aspect of the Poynting vector comes from the area (normal vector to a surface), not the power. >> Yes but what I was getting at is the 'normal' direction of the interacting vectors producing an orthogonal resultant. >> The Poynting vector is not energy; it is in units of power/area >> Yes I have concluded that aspect (acceleration squared). Normally i have seen the PV described as an energy vector..... that confused me for a while
Yes, it's what I call a "common usage error". Everyone says it, but it is not technically correct. As long as you know the details you are ok, but it is very easy to get tripped up. Another example is escape velocity. Everyone says "escape velocity" but it should really be "escape speed" or "escape energy" because the direction doesn't matter. As long as you don't actually hit the earth you will escape regardless of the direction you are heading provided your speed is high enough that your KE is greater than your PE measured from infinite separation. -Dale
">> "common usage error" Thanks DaleSpam I am glad I read it correctly... but only after I applied it and got pounced upon, then I got up again, LOL
