# Tag Info

## Hot answers tagged newtonian-mechanics

20

Great photo! Edit: My language is "sloppy" (I like talking physics in "lay person" terms so anybody can understand) but @dcmkee made really nice comment clarifying my answer for the more advanced people. Thanks @dcmkee! Since the plane is in a loop there is significant g's due to centripetal acceleration. The water was being accelerated upward$^{1}$ with ...

14

It doesn't actually have anything to do with the plane being upside down, or even changing from a vertical direction to a horizontal one. It's purely the vertical velocity that's at play here. Imagine water being thrown upward. You know what, imagine a fountain, a really big fountain. As soon as the water leaves the underground pump, it starts falling back ...

5

At the moment the picture was taken the plane, with mass $M$ was performing an inside loop, and was almost exactly inverted. It was moving at a speed $V$ in a vertical circle with radius $R$; both of these are chosen by the pilot as he execute the loop. The physics of circular motion requires that the plane experience a force towards the centre of the ...

4

The frictional force is opposing the tendency for the car to slide off the road. Think of a merry go round: If it is really fast you will struggle to stay on. If it goes really fast you will grab on to the bar and your body will point radially outward from the center of rotation. Eventually you be able to hang on and you will fly off. This is the same idea ...

4

No, these building are still tiny compared to earth's crust mass distribution. One would need to build whole mountain ranges to detect changes in earth gravity field with high precision instruments. And even those wouldn't changed earth orbit measurably because even a mountain range is tiny compared to the mass of the whole earth. However mountain ranges ...

4

The equation comes from Newton's second law: $$F = ma$$ Galileo didn't know calculus (because Newton and Leibniz hadn't discovered it yet) so he couldn't derive the equation mathematically. Since we do know calculus we know that acceleration is the variation of velocity with time: $$a = \frac{dv}{dt}$$ And also the gravitational force $F$ is equal to ...

3

Actual aircraft attitude (inverted with respect to the ground, in this case) is irrelevant. All that matters is that for the few moments long enough to pour the water and snap the picture, the aircraft is experiencing some positive g-load (pilot feels that he is pushed into his seat). The aircraft could be in a barrel roll or a loop. Either way, it is in ...

3

Your mistake is to believe that every action causes an equal and opposite reaction, in which case the reaction would cause a re-reaction, etc. Instead, the right way to think about Newton's Third Law is that whatever causes an action must simultaneously cause an equal and opposite reaction. The action/reaction don't cause each other; they are both caused ...

2

Surprisingly, the answer is that yes you do, though the effect is very small. To see this consider the following (highly exaggerated) diagram of the lift shaft: The Earth rotates at a constant angular velocity of one rotation every 24 hours ($\omega = 7.27 \times 10^{-5}$ radians/sec). The tangential velocity of a part of the lift shaft at a distance $r$ ...

2

Emf on a conducting object induces eddy currents. These in turn decay due to the electrical resistance of the object. What you end up with is energy in the form of heat. When you compare the two objects (essentially a conductor versus a non-conductor), a portion of the potential gravitational energy goes into generating eddy currents. That means the ...

2

Frictional force opposes sliding motion, basically. Car tires produce centripetal force by changing their angle relative to the rest of the car's orientation. The tires do not slide in the direction of the tires' orientation: they roll. Friction in this direction rotates the tires, or if the engine is applying force to the wheels during the turn, friction ...

2

Your equation is incorrect. The gravitational potential is $$\phi(r)=-GM\frac{3a^2-r^2}{2a^3}$$ when you're inside a uniform sphere of radius $a$ with total mass $M$. This is a quadratic potential in $r$, which is why it gives rise to harmonic exchange of energy when you oscillate between the planet surface and the core.

2

Let's suppose I have some system and I know $M$, the system's total mass, $\vec{r}_{cm}$, the system's center of mass position and $\vec{L}_{cm}$, the systems angular velocity in the frame where the center of mass is the origin. How do I find $\vec{L}'$, the angular momentum with respect to some other origin, say $\vec{r}_{cm} + \Delta \vec{r}$, which is ...

1

What you need is to match mechanical impedance between what the motor produces, and the tuned mass damper. Read more here http://www.bksv.com/doc/17-179.pdf and here http://arkansas.s.jniosh.go.jp/en/indu_hel/pdf/43-3-3.pdf

1

The value of initial velocity will be different for different angles θ with the horizontal.. So I got this result. $$u=(gR/(sinθcosθ-cos^{2}θ))^{1/2}$$ or $$u=(2gR/(sin2θ-2cos^{2}θ))^{1/2}$$ or $$u=(39.2/(sin2θ-2cos^{2}θ))^{1/2}$$ This is my attempt for the solution(i have attached image): From A to B displacement is FB From C to B displacement is ...

1

why wouldn't that straight line be in the direction of acceleration why do you think the acceleration line be in the direction of tangent? the tangent is where a body would have kept moving if the rope didn't pull it. so the vector of speed changes towards... where the rope is attached, i.e. perpendicularly. acceleration is the change in velocity ...

1

The fly would move out of the way. The fly (when right in front of the pilots face) has the same forward velocity as the plane. If the airplane accelerated forward then the fly would "fly backwards" (no pun intended) just as you think you do in a car that accelerates quickly. However, this is actually not you flying backwards, it is simply the car going ...

1

While I agree with the caveats made by dmckee in his comments, there is an obvious interpretation of stopping power as the change in momentum caused by the projectile. The mass and velocity of the projectile are $m$ and $v$ respectively, and the mass of the target is $M$. Since the target is stationary the initial momentum is just $mv$. Assuming the ...

1

When you solve a problem like this, you are using a system of reference (actually you use one in all problems, but here it is very explicit). In this case, the easiest one is y in the vertical and x in the horizontal. Almost all the forces are already in one of these 2 directions. Namely, you have all the weights pointing downwards, so in the -y direction ...

1

It will in general depend on the shape of the object. If it has a large concentration of mass at the edge you are lifting then the force will be close to its weight; if its mass is concentrated near the other edge then it will be very small. The general case is solved using the law of levers: If $d$ is the distance from the fulcrum to the centre of mass ...

1

If you are lifting on one edge and it is resting on the other edge, and the edges are an equal distance from the center of mass, then the answer is $$\boxed{F = \frac{1}{2} W}$$ If you are lifting with a distance of $\ell_1$ from the center, and the pivot is $\ell_2$ from the center then $$\boxed{ F = \frac{\ell_2}{\ell_1+\ell_2} W }$$ This is commonly ...

1

I am not sure what you meant by: "I figured I could simply calculate the magnitude of the components since that will give me the distance" But the idea is use the kinematics equations for x and y: $x(t)=x_{0}+v_{x0}t+1/2at^2$ and $y(t)=y_{0}+v_{y0}t+1/2at^2$ These equations are derived from integrating the acceleration function ...

1

"For every action, there is an equal and opposite reaction" This means, forces exist as pairs. When there is an interaction between $A$ and $B$, an action-reaction pair between them is produced. Which one is action and which one is reaction depends on your frame of reference. Now to the static friction. Your question is pretty vague so I'm going to ...

1

Suppose you have a satellite of mass $m$ at a distance $r$. If we assume the satellite is small enough to behave as a point mass the moment of inertia of the satellite is: $$I = m r^2$$ so its kinetic energy is: $$E = \tfrac{1}{2} I w^2 = \tfrac{1}{2} m r^2 \omega^2 \tag{1}$$ But for a body moving in a circle of radius $r$ at an angular velocity ...

1

First of all, mathematical definitions of force and momentum aren't really very intuitive or common-sensical. Just ask Aristotle for his common sense laws of forces! The fact that momentum is conserved in closed systems is a highly non-trivial fact, as is the Third Law. The reason that these laws exist at all is because you can't really 'see' or' feel' ...

1

I might be able to get you started a direction. Not necessarily the right one or a good one, but a direction. First, I chose a different place for $\theta=0$. It appears you chose it at the top of the loop, while I chose it on the right side. Oh well. Also, it's likely I have a typo somewhere... Newton's second law for this problem is ...

1

You can solve for $A$ and $B$ by using $r(0) = a$, $\theta(0) = 0$ and evaluating $\dot{r}$ and $\dot{\theta}$ in your solution and using the initial conditions for velocity. You will need an equation for velocity as a vector in polar coordinates. Furthermore, while you don't have $\theta(t)$ explicitly, it is a useful exercise to consider what happens to ...

1

Define the potential energy $$U(r) = -\int_{\infty}^r F(\bar r)\cdot(-d \bar r) = -\int_r^\infty F(\bar r)\cdot d \bar r$$ with the condition that it is zero at infinity. If the initial overall energy of the particle $E_0 = U(a) + \frac m2 V^2$ is larger than zero the particle will escape. (The only exception is the case that $k<0$ and the path is ...

1

A well executed barrel roll maintains the force balance you experience at rest with "gravity" oriented in the direction you experience as "down" (that is the direction from your head to your feet) due to centripetal acceleration. If you weren't looking outside, you might not realize the roll even took place (if the pilot is good). For those not convinced ...

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