How is acceleration calculated when only the direction changes and the magnitude remains the same? Guys, I know that my previous examples didn’t make sense and I’m sorry because I’m totally new to this topic and also I don’t think ya’ll actually understand my question. Acceleration is change in velocity over change in time,right? So, if the magnitude of velocity changes and the direction is same, acceleration is obviously given a value (for e.g. 2 $m/s ^2$ etc) and the direction of acceleration is the direction of change of magnitude of velocity. My question is, what if only the direction changes? What if there’s this super advanced object that can change its direction without changing the magnitude of velocity? How will the acceleration be calculated in that case? How can the magnitude of acceleration have value since there is no change in speed ( magnitude of velocity)? I’m also aware about Uniform Circular Motion and how acceleration changes constantly even when the magnitude of velocity doesn’t change and how it can be calculated. But uniform circular motion is not an one dimensional direction change. It will be of huge help if you could explain how acceleration is calculated when magnitude of velocity remains same and only direction of velocity changes in 1D motion (I.e., it goes in the opposite direction and for simplicity let’s assume that this body/object needs very little time to change its velocity from positive to negative) I’m still yet to learn calculus so if these calculations include calculus please do explain it a little in detail. Also, if an object can not change direction without changing speed in 1D motion, then please explain giving an example where the object takes very little time to change direction/ go back. I just want to know how the direction change is shown in acceleration. Thank you for baring with me and for your time.
123215
The formal definition of acceleration is      It is the measure of how fast the velocity changes with respect to time.   Now, when the magnitude of velocity changes, acceleration $$\vec{a} = \frac{d \vec{v}}{dt}$$ and its unit is $m.s^{-2}$ .  But what about the case where the velocity only changes its direction keeping its magnitude constant? The definition can then be      Acceleration is the measure of how fast the velocity changes its direction with respect to time.   But how can I calculate the acceleration then?? What should be the unit then??? Please help.
14637
While trying to understand the second law of Newton from "An Introduction to Mechanics" by Kleppner and Kolenkow, I came across the following lines that I don't understand:      "It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector. Although this turns out to be the case, it is not obviously true. For instance, if mass were different in different directions, acceleration would not be parallel to force and force and acceleration could not be related by a simple vector equation. Although the concept of mass having different values in different directions might sound absurd, it is not impossible. In fact, physicists have carried out very sensitive tests on this hypothesis, without finding any variation. So, we can treat mass as a scalar, i.e. a simple number, and write $\vec{F} = m\vec{a}$."   The lines above lead me to question:   Why is it not" obviously true" that force behaves like a vector?  Why is it not impossible for mass values to be different in different directions?
67918
I am studying angular momentum and I get the concept that it has a meaning or it is defined only with respect to a  certain point (often called as origin). I was looking at the realtionship between torque and angular momentum around an origin i.e.  net torque  = Rate of change of angular momentum (equation 11.29 in the picture). I think i get this relationship. Where I got confused when author adds a caution (full quote below) to the equation saying that:  If the center of mass of the system is not accelerating relative to an inertial frame, that origin can be any point.  However, if it is accelerating, then it must be the origin.  I am not sure if I get that statement and caution.  Not sure what the author means by the “origin” here.  Can't the origin be any point that we choose?   Let $\vec{\tau}_{\mathrm{net}}$ represent the net external torque, the vector sum of all external torques on all particles in the system.  Then we can write Eq. 11-28 as \begin{equation}   \vec{\tau}_{\mathrm{net}} = \frac{d \vec{L}} {dt} \qquad \text{(system of particles)} \tag{11-29} \end{equation} which is Newton's second law in angular form.  It says  The net external torque $\vec{\tau}_{\mathrm{net}}$ acting on a system of particles is equal to the time rate of change of the system's total angular momentum $\vec{L}$.  Eq. 11.29 is analogous to $\vec{F}_{\mathrm{net}} = d\vec{P}/dt$ (Eq. 9-27) but requires extra caution: Torques and the system's angular momentum must be measured relative to the same origin.  If the center of mass of the system is not accelerating relative to an inertial frame, that origin can be any point.  However, if it is accelerating, then it must be the origin.  For example, consider a wheel as the system of particles.  If it is rotating about an axis that is fixed relative to the ground, then the origin for applying Eq. 11-29 can be any point that is stationary relative to the ground. However, if it is rotating about an axis that is accelerating (such as when the wheel rolls down a ramp), then the origin can be only at its center of mass.
54306
Is time simply the rate of change?  If this is the case and time was created during the big bang would it be the case that the closer you get to the start of the big bang the "slower" things change until you essentially approach a static, unchanging entity at the beginning of creation?  Also, to put this definition in relation to  that "observers in motion relative to one another will measure different elapsed times for the same event." : Wouldn't it be the case that saying the difference in elapsed time is the same as saying the difference in the rate of change.  With this definition there is no point in describing the "flow" of time or the "direction" of time because time doesn't move forward but rather things simply change according to the laws of physics.   Edit: Adding clarification based on @neil's comments:  The beginning of the big bang would be very busy, but if time was then created if you go back to the very beginning it seems there is no time and there is only a static environment.  So it seems to me that saying time has a direction makes no sense.  There is no direction in which time flows.  There is no time; unless time is defined as change.  So we have our three dimensional objects: and then we have those objects interact.  The interaction is what we experience as time.  Is this correct or is time more complicated than this?
52873
thanks for any help.  I'm trying to show that in a 2body problem, angular momentum is conserved given that $\dfrac{dp}{dt}=\dfrac{-GMm(rv)}{r³}$, where p is momentum, t time, G gravitational constant, M mass of 1 object, m mass of the other, (rv) is the vector between them and r is the magnitude of the vector between them.  I've had a lot of attempts but I don't seem to get very far.  Where L is angular momentum, $r_1$ is position vector of mass M and $r_2$ is position vector of mass m, $p_1$ is momentum of mass M and $p_2$ is momentum of mass m.  L=$r_1Xp_1+r_2Xp_2$  If $\dfrac{dL}{dt}=0$ then dL will be conserved over time, applying product rule             $L=\dfrac{dr_1}{dt}Xp_1+r_1X\dfrac{dp_1}{dt}+\dfrac{dr_2}{dt}Xp_2+r_2X\dfrac{dp_2}{dt}$  noticing that $p_1=m\dfrac{dr_1}{dt}$ and $p_2=m\dfrac{dr_2}{dt}$ clearly $p_1$ is parallel to$ \dfrac{dr_1}{dt}$ and $p_2$ is parallel to $\dfrac{dr_2}{dt}$ and hence when cross producted with them, the result is 0.  Giving $L=r_1X\dfrac{dp_1}{dt}+r_2X\dfrac{dp_2}{dt}$  I gather I must now show that either these two terms are 0, or they are equal and     opposite. I can't see any reason why they would be 0, and (slightly guessing) it seems to    make sense to me that they are terms related to how the angular momentum of one body changes    with the other body, and I assume they must therefore be equal and opposite. However when I    do  $r_1X\dfrac{dp_1}{dt}=-r_2X\dfrac{dp_2}{dt}$      I notice $\dfrac{dp_1}{dt}=\dfrac{-GMm(rv)}{r³}$ and $\dfrac{dp_2}{dt}=\dfrac{GMm(rv)}{r³}$ (since (rv)'s direction is flipped)     Therefore $r_1X\dfrac{-GMm(rv)}{r³}=-\dfrac{r_2XGMm(rv)}{r³}$     Defining $w=\dfrac{GMm}{r³}$:  $r_1X(rv)(-w)=r_2X(rv)(-w)$     Scalar products are commutative so the (-w)'s easily cancel giving  $r_1X(rv)=r_2X(rv)$  Which then the only solutions are if all components of $r_1$ are equal to all components of      $r_2$, however angular momentum should clearly be conserved in all positions, not just when the masses are ontop of each other.  Thanks again for any help.
64188
A while back in my Dynamics &amp; Relativity lectures my lecturer mentioned that an object need not be accelerating relative to anything - he said it makes sense for an object to just be accelerating. Now, to me (and to my supervisor for this course), this sounds a little weird. An object's velocity is relative to the frame you're observing it in/from (right?), so where does this 'relativeness' go when we differentiate?   I am pretty sure that I'm just confused here or that I've somehow misheard/misunderstood the lecturer, so can someone please explain this to me.
5257
As a layman, i can calculate approx "displacement" just by observing the moving object. And accurately by using a simple "scale". Similarly, again, I can calculate angle from origin by using displacement in $x$ and $y$ dimensions. Similarly I can use a stopwatch and scale to understand velocity.  But when I read about , first of all it confuses with the english word "force", we use in real life. To some extent I am sure, it has nothing to do with that. So exactly what is it in the real sense. How can a layman, see force as, just like he can see displacement or angle? Or force is just a quantity defined by physicists to simplify the combinations of $ma$, they might be facing every time. And thus came up with term "force" ( which is similar in spelling to english word in oxford dictionary). And lastly, why not force has just been called as something proportional to mass and it's displacement or velocity. Why something at the level of change of velocity has been used to define it.
92255
Referring to unidimensional motion, it is obvious that it doesn't always make sense to write the speed as a function of position. Seems to me that this is a necessary condition to derive formulas like:  $$v^2=v_0 ^2 +2\int_{x_0}^{x}a\cdot dx$$  In fact, in the first step of the demonstration (the one I saw, but I think that this step is crucial) it's required to write $a=dv/dt=(dv/dx)(dx/dt)$, that doesn't make sense if $v$ isn't a function of $x$.   When can one rigorously write $v=v(x)$?
3242
We have the image below displaying the uniform velocity by time-distance graph. At every point velocity is constant but what if distance and time both become zero as at origin in the graph is? The velocity must become undefined as $\frac{0}{0}$ is undefined in mathematics or shall we call it zero? If zero then why?
26988
Suppose you have an object with zero for the value of all the derivatives of position. In order to get the object moving you would need to increase the value of the velocity from zero to some finite value. A change is velocity is acceleration, so the value of the acceleration would have to increase from zero to some value. You would also need to increase the jerk from zero to some value to have a change in acceleration.  Is there an infinite series of higher derivatives of position for this to work? Or am I missing something?
99658
I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.  One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).  So my question would be the following: How do these two phenomenons describe the same law of motion?   PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?
71958
We know that it takes no energy to change the direction of a vector, we know also that it takes no energy to displace a body in motion if a push is applied exactly at a right angle on its center of mass.  Considering that, if that is true, can you explain why a simple change of direction without any increase of speed is considered an acceleration?
68857
I was confused with this equation like why force is equal to $ma$ and not $mv$? Then I read the same question here on stack exchange, it's answers almost made sense but I still can't visualise the situation where a thing is moving at a constant velocity, would it then have no force? But if I think this scenario in my head, it would definitely have force when it hit something. So I wonder if I am mis-associating the terms in the equation?
85569
Why is the radians implicitly cancelled?  Somehow, the feet just trumps the numerator unit. For all other cases, you need to introduce the unit conversion fraction, and cancel explicitly.  Is it because radians and angles have no relevance to linear speed (v), so they are simply discarded?
105833
I recently watched  by Veritasium where he talks about the One Way Speed of Light and talks about the limiting case where in one direction the speed of light is $c/2$ while it's instantaneous in the other. He also says this is perfectly fine according to our Physics theories. He also points at Einstein's assumption in his famous 1905 paper where he assumes that the speed of light is same in all directions. This made me ask this question is taking the speed of light same in all directions an axiom of some sort? As I've often read no information can be sent at more than the speed of light but here one-way taking the speed to infinite makes no difference. So are all of our physics theories based on the assumption and what would happen if light turns out to be moving at different speeds in different direction? Will that enable transfer of information faster than the speed of light and is there any way for us knowing that the transfer happens faster than the speed of light? The video takes a Earth Mars case where he says it isn't possible for us to every realize this discrepancy but is there a more general proof which says it isn't possible
103728
Consider this situation:  A ball is moving forward and undergoing rotation. Assume that it is not slipping. Eventually, the velocity and rate of rotation of the ball decrease, and it comes to a halt. But if you observe the direction of friction (when the ball is rotating clockwise), you will see that the friction should have provided a clockwise torque to the ball and the angular velocity of the body should have increased. But this doesn't happen. Why?
6669
I have a background in calculus but don't really know anything about physics.  Forgive me if this is a really basic question.  The equation for distance of an accelerating object with constant acceleration is:  $$d=ut        +\frac{1}{2}at^2$$  which can also be expressed  $$d=\frac{\mathrm{d}x}{\mathrm{d}t}t+\frac{\mathrm{d^2}x}{\mathrm{d}t^2}\frac{t^2}{2}$$  (where x(t) is the position of the object at time t)  That's fine for a canonball or something like that, but what about a car accelerating from 0 to cruising speed?  The acceleration is obviously not constant, but what about the change in acceleration?  Is it constant?  I suspect not.  And then what about the change in the change of acceleration, etc. etc.?  In other words, how does one know how many additional terms to add in the series?  $$d=\frac{\mathrm{d}x}{\mathrm{d}t}t+\frac{\mathrm{d^2}x}{\mathrm{d}t^2}\frac{t^2}{2}+\frac{\mathrm{d^3}x}{\mathrm{d}t^3}\frac{t^3}{3}+\frac{\mathrm{d^4}x}{\mathrm{d}t^4}\frac{t^4}{4}\cdot etc. \cdot ?$$