# 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 ...

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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 ...

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 ...

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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 ...

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 ... 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 The roll doesn't have to be in a fixed place for a torque to be generated. Torque will be generated regardless of a fixed axis or not and your toilet paper will unroll eventually. Torque is generated because as @CuriousOne said above, the force is acting somewhere other than through the centre of mass thus creating an unbalanced force on one side which ... 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 ... 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 ... 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 The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ... 8 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 ... 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 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 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 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 $$You already have the mass, and you can Google for the ... 6 Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other. If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. ... 6 Electrons in a conducting disk in order to maintain equilibrium will have to have a centripetal force on them equal to the local change in potential energy with respect to a change in radius, that is$$ 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 ...

5

Altair, Vega, and Regulus A are perhaps the most famous examples of stars that have been "flattened" by rapid rotation. Some studies (mentioned in Yoon et al. (2010) suggest that Vega is rotating at 70-90% percent of the speed at which it would break up (its rotational velocity is about 20 km/s). Regulus is even closer to this breakup speed: If its ...

5

A rigid body can not in general be modelled as a mass point. This is possible in celestial mechanics, as the forces encountered there act uniformly and therefore can be effectively described as forces acting on the centre of mass. In general, you have to consider the orientation of the body as well, then one gets the equations of motion for the centre of ...

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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 ...

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John Rennie's answer is correct in the special case that angular momentum and angular velocity are parallel to one another. This is not always the case as moment of inertia is a second order tensor. Angular momentum is given by $\boldsymbol L = \boldsymbol{\mathsf{I}} \boldsymbol \omega$. Differentiating this gives the rotational analog of Newton's second ...

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There is a general and simple formula to calculate the moment of inertia with respect to some axis if the moment of inertia with respect to another axis is known. I am pretty sure you'll be able to find it in your textbook. The theorem is parallel axis theorem.

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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 ...

5

I think you have misunderstood how friction works here. The friction you have written down is (typically) the minimum friction needed for the 'non slip' to occur. Imagine a very faint slope. You will only need a small amount of friction to avoid slipping. As you make the slope steeper more friction is needed for the 'non slip'. Eventually the friction will ...

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