# Tag Info

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The astronaut can change his or her orientation in the same way that a cat does so whilst falling through the air. After the transformation, the astronaut is still and angular momentum is conserved. There is a rather beautiful way of understanding this rotation as an anholonomy i.e. a nontrivial transformation wrought by the parallel transport of the cat's ...

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For those that are cat-challenged, here's an alternative explanation and demonstration you can try at home! This demonstration was taught to me by my math lecturer. All you will need is: A swivel chair and a heavy object (e.g. a big textbook) Stand on the seat of the chair (watch your balance now) holding the heavy object. Extend your arms forward ...

6

Suppose you pick two people at random. From one, you pluck a single hair from their head. Is it possible to tell who had the hair plucked by weighing the people? Technically, plucking a hair makes a person very slightly lighter, so you get a tiny bit of information about who had the hair plucked by weighing the people. But the information is very slight ...

4

That's exactly the case. If you look at the trajectory of any given spacecraft, you will see that it has a few burns of the rocket engines punctuating very long periods just coasting along in orbit around some other body. For example, the flight path of Apollo 8 has something like eight different rocket burns: launch, translunar and transearth injection (to ...

4

If the ladder is slipping on the floor as well as the wall, then the point of rotation is where the two normal forces intersect. This comes from the fact that reaction forces must pass through the instant center of motion, or they would do work. In the diagram below forces are red and velocities blue. If the ladder rotated by any other point other than S ...

4

You're confusing the acceleration of your car with the acceleration in a collision. You actually have to look at it "backwards" from what you've described above. That is, in the collision you don't do a $F = ma$ calculation where $a$ is the acceleration of your gas pedal. Instead in the collision you have a force $F$ resulting from the collision and you ...

3

Look on the water from the point of view of the accelerated reference frame oriented in such way that the surface of the water is parallel to plane $x'y'$ and depth below the water surface is measured by $z'$. In this frame, the total gravity (due to Earth's gravity and due to inertial force of acceleration) is directed perpendicular to the water surface and ...

3

You are right to be wary. When the gravitational field is nonuniform, the centre of mass and the centre of gravity are in general different. The centre of gravity is the point around which the nett torque from the gravitational forces is nought. In your problem, where gravity varies with position $\vec{r}$, we seek the position $\vec{r}_0$ where the nett ...

3

Another way to think about Newton's second law (and the way he originally defined it) is $F=\dfrac{d\rho}{dt}$, where $\rho=mv$ is momentum and $\dfrac{d\rho}{dt}$ is the rate of change of momentum. I think you meant to say that the obstacle will exert a force on you - and that is correct. If you could calculate your change in velocity, and the amount of ...

3

The ladder falls because it experiences unequal moments from the normal reactions at both its ends. That is to say that the surface pushes on the ladder from the bottom as well as the side. In the absence of tangential contact forces such as friction, the ladder rotates and falls. To solve a problem with such a situation, you may choose any point as the ...

3

There's another way to do this also, more akin to how spacecraft actually do it: Take a weight on a string, hold it up and spin it. You'll turn in the opposite direction. When you stop it you also stop turning. Of course this will produce an off-axis force that will be a real pain to deal with. Real spacecraft do it by means of a set of internal wheels ...

3

Other answers have pointed out other ways that might be more efficient, but one very simple way to do it is as follows: start with both arms parallel to the body. Then swing them both backward, up over the head, and then back down in front of the body, leaving them back in the starting position. After this manoeuvre, the body will be oriented in a slightly ...

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Actually, to the best of my knowledge, your solution is correct. Your logic is also very sound. The displacement is calculated from your formula. Note how the guys that send people to orbit also agree with you. And also, I'm not sure about what is the policy regarding Wikipedia links, but take a look at this article. More specifically, the following quote: ...

2

When a disk or other object is rotating on a horizontal surface with constant velocity, there is no static frictional force. Your logic is correct: if there were a horizontal force, the center of mass would be accelerating. If the rolling object suddenly encounters a frictionless surface, it would continue to satisfy the rotating without slipping condition. ...

2

Yes. Your analysis is completely true on the basis of Newtonian Mechanics. But upon observation of the universe this turns out to be wrong, as once the objects in consideration approach the speed of light, we have to apply Relativistic Mechanics. So, as far as Newtonian Mechanics is concerned, objects can have infinite velocity and momentum. But they can't ...

2

I suspect that you may be under the mistaken impression that there is no gravity in space. This is a common belief since we all can see the astronauts floating in "zero g" when on the ISS are some other spacecraft. However, we can easily dispense with this misconception by asking "what keeps the ISS in orbit around the Earth if there is no gravity?". Of ...

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In the situation at hand, you'll never be able to achieve uniform circular motion. $\frac{d\vec{v}}{dt} = \frac{1}{m}\sum \vec{F}_{\text{ext}}$ This is a vectorial equation. If you look at the picture you've drawn, you have forces on the radial as well as the tangential direction. On the radial direction, there is the tension force, and $mg\cos\theta$. ...

2

Those equations are rather tautological. The gist is that it doesn't matter whether you have some acceleration or gravitational attraction, they're really indistinguishable. So you basically get a $g' \equiv g + a$ as new "compound down acceleration". Then you write down the exact same equations as for the earth-rest-frame case, but with this modified $g'$ ...

2

User Sahil Chadha has already answered the question, but here's the math and a pretty picture for anyone who is unconvinced that you're right. Since the train is accelerating, from the perspective of an observer on the train, the ball will experience a (fictitious) force in the direction opposite the train's travel having magnitude $ma$ where $m$ is the ...

1

Fluid dynamics problems such as this are generally best approached by control volume analysis. Consideration of conservation of mass, momentum, energy, and sometimes angular momentum for an isolated control volume system generally provide an engineering answer. To figure out the force exerted on the pipe by the fluid it would seem appealing to isolate the ...

1

Since you are dealing with an inelastic collision, energy is not conserved when the bullet hits the block. You should try to find a relation between the initial velocity of the bullet and the velocity of the combined system (bullet+block) after the collision from conservation of momentum.

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You do not give any indication of your physics background so I will answer in simple terms. Are they saying, for example, that a source which has a flux emanating from it can be treated as a point particle, even if it's spherical? How can spatially extended objects behave like point particles? A simple example is the solar system. To first order ...

1

The stretching of the air-liquid meniscus matches the pressure exerted by the liquid, not the mass of liquid in the tube. In a closed tube of water the pressure at the top of the tube is $P = P_0 - \rho g h$ where $h$ is the vertical distance (P_0 is atmospheric pressure), so in the diagram above $P_1 = P_2$. So if we replace the closed end of the tubes ...

1

Let's say you roll a ball (of mass $m$) down an inclined plane of angle of inclination $\theta$ and coefficient of static friction $\mu_{static}$. Then you know a force parallel to the inclined plane acts on the ball through its center of mass. Another force parallel to the surface acts in the opposite direction of motion as follows, The force $\vec F = ... 1 Parallel axis theorem does not care for mass distribution along the rotation axis, only away from the rotation axis. The answer is yes then. Below is the formula for the total mass moment of inertia of the disk +$N$masses, each attached a distance$r_i$from the axis of rotation. $$I_{total} = I_{disk} + \sum_{i=1}^N \left( I_i + m_i r_i^2 \right)$$ 1 You're getting confused about normal reaction. The normal reaction is a "reactive" force, so it depends on the actual active force. When you're taking a sharp turn, centrifugal force pushes your car outwards. Since the center of mass of the car is above the level of the wheels, it tends to create a moment which pushes the outer wheels down and the inner ... 1 The ratio of large-to-small piston diameter is$1.8:0.045$or$40:1$The ratio squared is$1600:1$Since volume depends on diameter (or radius) squared, and the fluid volume is constant, the piston motion must scale as$1:1600$Input plunger moves$1.5\text{ m}$, so output moves$1.5/1600$or$0.0009375\text{ m}$1 Great question; I remember being so confused by this when I first took analytic mechanics. The components of the angular velocity "in the body frame" aren't zero because when one writes these components, one isn't referring to measurements of the motions of the particles in the body frame (because, of course, the particles are stationary in this frame). ... 1 In the cases where you have static friction, the forces will always be defined by the looking at the system and applying the constraints(in other words$F_s\le \mu N$will only give an upper bound). On the other hand when you are dealing with kinetic friction, it can be easily derived from the famous$F_k=\mu N\$. As an example, let's solve this problem(As ...

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