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145

The overshoot behavior you noticed is called cogging and occurs when the magnet arrangement in the motor "catches" the rotating magnetic core of the motor during shutdown and jerks it back to one of the local strong spots in the field. You can demonstrate this yourself by carefully rotating the fan blade around with your finger when the motor is ...


25

The ball will slow down to terminal velocity. This is because the force of air drag increases with increasing speed. Terminal velocity is the speed where the force of air drag equals the force of gravity, so the total force is zero and the object travels at a constant speed. If the ball has a higher speed, then it will have an air drag force greater than ...


6

Your PC is a box with limited vents for air to enter/exit. Especially if these clog up with dust, the fan could create a noticeable pressure differential between inside and outside of the box. After the fan turns off, that equalizes, forcing some air backward through the fan, and causing it to rotate backward. Maybe. Another possibly, as pointed out in ...


5

Well, few problems here, which all seem to come down to a disregard for units of measurement. First, the quantities you take to be equal to $1$ in the beginning still have dimensions (apart from the drag coefficient, let's keep that at $1$ for simplicity), so they're not just "$1$". In fact, you see that shortly after this, you're writting results ...


4

Terminal velocity is achieved when the drag force $f=Dv^2$ (where $D$ is a constant and $v$ is the object's speed) equals the gravitational force $F_g=mg$. Equating these forces is the condition for terminal speed, which is $\displaystyle v_T=\sqrt{\dfrac{mg}{D}}$. If the ball is thrown with an initial velocity greater than the terminal velocity, then $f>...


4

The differential equation A general trick in these cases is to solve the differential equation first for the speed, which is easy and first-order, and then integrate the speed to get the position. Indeed, this is a differential equation of the form $$d^2x/dt^2=A-B(dx/dt)^2$$ If we write $v=dx/dt$ this becomes $$dv/dt = A-Bv^2$$ which can be solved despite in ...


4

Did the fan really rotate backwards, or merely appear to? You may be looking at a Wagon-Wheel Effect. Basically, when a wheel or propeller rotates at a certain speed, particularly if you're using fluorescent lighting, which has a natural flicker to it that isn't always obvious, the object can appear to stop or rotate in the opposite direction. All the ...


4

On balance, I think @nielsneilsen's answer is likely the correct one. But there is another possibility. Is there a control loop in play? Almost universally, practical control systems for hitting a desired position, velocity, flow rate or whatever use a PID control loop, sometimes omitting the P or D element depending on the application. Other control ...


2

I also think the effect is cogging. This torque is position dependent and its periodicity per revolution depends on the number of magnetic poles and the number of teeth on the stator. Cogging torque is an undesirable component for the operation of such a motor. It is especially prominent at lower speeds, with the symptom of jerkiness. Cogging torque results ...


2

The equation $v^2=u^2+2as$ only applies to an object that is accelerating at a constant acceleration $a$. Once you take into account drag, the acceleration of a falling object is not constant - it starts at $g$ when $t=v=0$, but then decrease asymptotically towards $0$. So you cannot use $v^2=u^2+2as$. A falling object with drag never actually reaches its ...


2

From the wikipedia article about Javelin throw In 1986 the technical committee of the International Association of Athletics Federations introduced new specification for the standard competition javelin. Among other things it was specified that the center of mass should lie forward of the center of air resistance. This new specification gives a "nose ...


1

If I understood correctly your notation, we are dealing with and ordinary differential equation here: $$ A\ddot{x}=B - C\dot{x}^2 $$ If we define function $y=\dot{x}$, it becomes $$ A\dot{y} = B -Cy^2, $$ which is trivially solvable $$ \frac{Ady}{B - Cy^2}=dt $$ and then integrating both sides. Note that this is a particular case of the Ricatti differential ...


1

Throwing a spear straight without its tip falling or rising and hitting a target with it requires that the spear possess no angular velocity about its center of mass when released. This is a nontrivial task that takes a lot of practice. The spear thrower has to move his or her arm through an shallow arc while throwing the spear. To make the spear fly ...


1

There may be no initial torque, but the spear cleaves the air to pass through. Air is composed of molecules that act with Newton's third law. Each molecule displaced contributes to the friction the spear has on its path. As a completely symmetric spear is very difficult to achieve, and unnecessary for the function of a spear, torque appears.


1

I can't exactly speak for the derivation as you have laid it out here because it doesn't really make sense... perhaps there is additional text that sheds light on what the set of equations means. However, let's look at what the thrust equation is saying. Newton's second law is often expressed as $F = ma$, but in reality Newton's second law says that the time ...


1

Hi Reading this is interesting. The reason that the fan goes backwards is simple. It is a Direct Current motor which has a capacitor across the power terminals to keep the power factor near to unity. A motor is a wound component which will produce a lagging current so a capacitor is added to bring that current nearer to unity (A capacitor creates leading ...


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