Force and Velocity directions

Question

When ever the force and velocity are in thesame direction(0°), an object speeds up. When ever the force and velocity are in opposite directions(180°), the object slows down. Also , when ever force and velocity are at 90° to each other, the object is known to describe a circular path with constant velocity. Now what happens if Force and Velocity are at some angle different from 0°, 90° and 180°.

Answer 1

The concept needed here is about the components of a vector quantity (in your case the velocity and the force) . A vector quantity can be projected along any of the perpendicular axis (X , Y or Z axis).

For convenience take the velocity to be along the X axis. And the force $F$ to be in the first quadrant of the X-Y plane at some angle $\theta$ with the X axis. This force will have some components along the X axis I.e.($F \cos \theta$) as well as along the Y axis I.e. ($F \sin \theta$). The component along the X axis will increase the speed while the component along the Y axis will make it follow a curved path (as you said in your question).

The same concept can be applied for any direction of the force.

Hope it helps ☺️.

Answer 2

An instantaneous mixture of both. No, really!

I can give you the boring answer which is “literally anything else.” Any alternative to “keeps going in a straight line while changing speed” or “goes in a circle at constant speed” must have some other description. And like that’s fine, I guess. But it doesn’t help you do physics.

What will help you do physics: If you have a velocity vector $\vec v$ and an acceleration vector $\vec a$ and they are at some angle $\theta$ then the speed will increase (or decrease if the cosine is negative!) by an amount $\|\vec a\| \cos\theta$ per unit time, where $\|\vec a\|$ is the overall magnitude of acceleration. If the velocity and acceleration are changing, then this is only a good approximation for very short times, much shorter than they typically change. That's why I say “instantaneously,” I want you to think of a very very short period of time that we could call an instant.

Furthermore it will curve over this instant as if it is following a circle with radius $\|\vec v\|^2/(\|\vec a\|\sin\theta)$ . Of course if $\sin\theta=0$ you are dividing by zero and the proper way to think about this is a circle of infinite radius centered on infinity, which is a straight line. On the other hand $v=0$ requires some actual thought, the circle has radio zero so what does that mean? Basically it means that it travels in a straight line in whatever direction it accelerates, there is no curvature to stop it. The curvature is, in some sense, inertia, if that makes sense. The more momentum you have going one way, the less some force is going to curve the trajectory, conversely if you have very little momentum the force immediately pulls you in whatever direction it wants because you have no momentum to resist its bidding.

We call these projections $\|\vec a\|\cos\theta$ and $\|\vec a\|\sin\theta$ the parallel and perpendicular components of the acceleration, and for any trajectory that is more complicated, we really do split it into instants in each instant can be described with a change in speed and a little segment of a curve.

Answer 3

Remember that $\frac{\vec{dv}}{dt}=\frac{\vec{F}}{m}$, which implies $\vec{v}(t+dt)=\vec{v(t)}+\frac{\vec{F}}{m}dt$, for small change in time $dt$.

The acceleration vector $\frac{\vec{F}}{m}$ can be written as a sum of two vectors, one parallel to the direction of $\vec{v(t)}$ and the other perpendicular to it. To split $\vec{F}$ into two vectors like this, you can use the dot product of $\vec{F}$ with $\vec{v}$. Let's call these two vectors $\vec{a_1}$ and $\vec{a_2}$.

So $\vec{v}(t+dt)=\vec{v(t)}+\vec{a_1}dt+\vec{a_2}dt$

When the angle is 90 degrees, $\vec{a_1}$ is zero, when it's zero degrees, $\vec{a_2}$ is zero. Otherwise, both terms are present.

Answer 4

Simple, it will do both i.e. it will change direction and speed and that could be calculated using vectors.

Answer 5

The power provided by a force is computed by the vector dot product with the velocity vector

$$ P = \vec{F} \cdot \vec{v} $$

In terms of magnitudes, the above is

$$ P = \| \vec{F} \| \| \vec{v} \| \cos \theta $$

where $\theta$ is the angle between the force and the velocity vectors. As you can see when $\theta =0 $ then $\cos \theta = 1$ and the force provides power (increases energy). When $\theta = 180 \deg$ then $\cos \theta = -1$ and the force removes power (decreases energy). Calculate the cosine of the angle to get all the in between cases.