# Tag Info

90

There are plenty of satellite galaxies orbiting larger galaxies. The question is how long are you willing to wait for an orbit? The Milky Way has a mass $M$ of something like $6\times10^{11}$ solar masses, or $10^{42}\ \mathrm{kg}$. The small Magellanic Cloud is at a distance $R$ of $2\times10^5$ light years, or $2\times10^{21}\ \mathrm{m}$. A test mass ...

47

I'm going to assume the bottom end of the rod is fixed, so the rod rotates around it. I think this is what you have in mind - shout if it isn't. So at some point during its fall the rod looks like: The mass of the rod is $m$ and the mass oif the weight on the end is $M$, and I've drawn in the forces due to gravity. To write down the equation of motion ...

41

It depends on where on Mars you toss the coin, and how high you toss it. In a rotating frame of reference, an object in motion appears to be affected by a pair of fictitious forces - the centrifugal force, and the Coriolis force. Their magnitude is given by $$\mathbf{\vec{F_{centrifugal}}}=m\mathbf{\vec\omega\times(\vec\omega\times\vec{r})}\\ ... 31 It is because the moment of inertia is not a conserved quantity. The statement that an isolated body can't change its position is more precisely the statement that an isolated body cannot change the position of its centre of mass. The position of the centre of mass, {\bf R}, is given by:$$ {\bf R} = \frac{1}{M}\sum m_i {\bf r}_i $$where M is the ... 27 They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally ... 25 As many others point out, there is friction present, otherwise the wheel wouldn't grap the surface and pull the car forward. But you are talking about a different kind of friction. There is a possibility of different kinds of friction: Kinetic friction, if the wheel ever slides and skids over the asphalt. This is friction between objects that slide over ... 22 In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ... 17 Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately:$$ P = ...

14

You are right. To open the door during the same time interval (for pushing at $a$ and $b$), you should induce the same angular acceleration. Since rotation in both cases is about the same axis, this means you need the same torque, this gives $$\frac{F_a}{F_b} = \frac{r_b}{r_a}$$ where $r$ is the distance from the point of contact to the axis. However the ...

12

How can we detect Earth's spin? Apparent motion of Sun You will have observed that the sun reappears every 24 hours. There are two common explanations for this. One of them is that the earth rotates with a period of approximately 24 hours - this is the only explanation supported by the scientific evidence. The main alternative had a rather convoluted ...

11

What does this small change means in form of Rotational Kinetic Energy? There's a problem with your calculation: You assumed a constant value for the Earth's moment of inertia. The Moon and Sun raise tides on the Earth itself. These Earth tides result in subtle changes in the Earth's moment of inertia. The signature of these tides can easily be seen in ...

11

This apparent paradox is actually not a paradox at all. Infact, it is because of Newton's second law, that we can say that the object will rotate exactly about it's centre of mass. This is true for any body on which net external force is 0, but net torque is present. In Newtonian mechanics, centre of mass serves to simplify calculations, for exactly the ...

11

John correctly stated that this is possible because re-configuring our bodies allows us to change our moment of inertia, but not our mass. As the question was about an intuitive explanation, consider adding a series of floating weights to get an analogous situation for translational motion: The astronaut stretches their arms above the head, grabs a weight, ...

11

anyway, how likely is it the ice ages could be explained by the earth 'realigning' so that polar regions would migrate over the surface of the earth? How about zero? The geological evidence of the Ice Ages clearly says that, between the ice episodes, the ice did not move. It's just that the polar caps shrank. For instance, the extent of the last ice ...

10

The physics 101 answer is no: it takes more force, but it is compensated by the smaller displacement so the energy stays the same. If we start with a static door, and we end up with a door rotating at some speed, the energy into the door is the work done by the force and it must be the same independently from the point where the force was applied But let's ...

9

Ernie is close to the correct answer, but the fundamental thing that needs to be considered is how the internal energy of the body flows. I researched this in a very interesting book I''m still reading, Principles of Animal Locomotion . Chapter 7 addresses running and section 7.5 discusses Internal Kinetic Energy . Limb accelerations can store kinetic energy ...

9

The coin will come back to your hand just like it would on the earth. The effect of atmosphere is negligible comparing to the coin's inertia, so the horizontal position of the coin relative to your hand will hardly be affected. The rareness of the atmosphere will only affect the vertical motion of the coin, like how quickly the coin will fall into your hand. ...

9

Yes, for the simple reason that you're not tossing the coin very high (presumably, anyway). You seem to think that on Earth, atmospheric drag is what keeps the coin "glued" to the tossing frame of reference, but that isn't really a factor at all. Say that you're on Earth, at sea level, on the equator, and you toss the coin 3 meters straight up. Neglecting ...

7

I assume that you're imagining two sticks whose bases stay stuck still to the ground as they tip over. In this case, what we should do to calculate the tipping rate (I wouldn't exactly say "falling"-- I think it's misleading) is consider the torque applied to each stick as it tips. What the answer really comes down to is the struggle between torque (the ...

6

Speculating here... I suspect that for light weights the answer is yes - with the right technique. Your center of mass moves up and down which requires energy being absorbed and expended by your legs. Moving your arms with small weights should allow you to even out the motion, lowering the peak stress on your legs so they tire more slowly. In a sense you ...

6

Electrons in a conducting disk in order to maintain equilibrium will have to have a centripetal force on them equal to the local change in potential energy with respect to a change in radius, that is $$m_e\omega^2 r = -e{d\phi\over dr}$$ After integrating, we get a potential difference between the center and a point R out $$\Delta\phi = -{m_e\omega^2 ... 6 Your intuition is correct. For the ball to change its angular momentum (to go from "backspin" to "forward spin"), there needs to be a net torque acting. There are two forces on the ball: gravity, and the normal force of the slope. Both these forces act through the center of mass - so neither force adds torque. Without torque, there is no change in angular ... 6 Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other. If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. ... 6 Even if I ignore wind and the drag forces and only consider the rotation of the earth the bullet will not hit the ground at the same place from where it was projected. There will be Coriolis effect. Coriolis effect: The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame. I suggest this ... 6 As pointed out by lemon, two angles are enough to specify a direction in a three dimensional coordinate system, but another is needed to specify a complete coordinate transformation. You can think of a rotation transformation in three dimensions as a mapping between two different coordinate systems. Two angles are needed to specify the relative pointing ... 6 No ,the centre of mass would not accelerate,the object will rotate about the centre of mass. If the object is on a table(or anywhere here on the earth) and if you apply the forces as shown in the fig it may not be rotated about the centre of mass because of the frictional forces acting on the object,but in space where there is no friction the object will ... 6 The rotational energy of a body is given by:$$ E = \tfrac{1}{2}I\omega^2 $$where I is the moment of inertia and \omega is the angular velocity. For a uniform sphere the moment of inertia is related to the mass of the sphere, m, and the radius of the sphere, r, by:$$ I = \frac{2}{5}mr^2 $$You already have the mass, and you can Google for the ... 6 No, in general they do not. You can work this out from the geometry of Ackermann steering, discussed on my website article "Parallel Parking a Car". In summary: look at the defining geometry for Ackermann steering, which I have sketched below: Ackermann steering is defined by the intersection of the central unit normals to (axes of rotational symmetry ... 6 The dynamics of a ball rolling down an incline is interesting. Let's start by figuring out the forces that come into play for the non-slipping case (mass m, radius R, angle of ramp \theta): If we consider the motion of the ball as a rotation about point P, then the torque is given by$$\Gamma = mgR\sin\theta and the moment of inertia about $P$ is ...

5

To a first-order effect, there would be no change. But one consequence of melting is that the water moves to other places. Water that moves from the poles to other areas on the surface of the earth would serve to (slightly) increase the moment of inertia of the planet. This is because the mass of the water would be farther from the rotational axis. The ...

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