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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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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 $$\mathbf{\vec{F_{centrifugal}}}=m\mathbf{\vec\omega\times(\vec\omega\times\vec{r})}\\ ... 27 They do! There's an entire class of galaxy, called a 'satellite galaxy' which is defined entirely based on them orbiting a larger galaxy (which would be called a 'central galaxy'). Our own milky-way is known to have many orbiting satellite galaxies, or at least 'dwarf-galaxies'. If dwarf-galaxies aren't enough, the milky-way itself is gravitationally ... 25 As many others point out, there is friction present, otherwise the wheel wouldn't grap the surface and pull the car forward. But you are talking about a different kind of friction. There is a possibility of different kinds of friction: Kinetic friction, if the wheel ever slides and skids over the asphalt. This is friction between objects that slide over ... 23 In this case, gravity is still an external force. In a zero-g environment, the mouse would also begin to move around the inside of the wheel, opposite the rotation it causes in the wheel, which would keep the angular momentum at zero. This would happen because the only way for the mouse to exert a force on the wheel and rotate it is for it to push itself in ... 21 The fan motor provides a torque \tau which has to accelerate \alpha the fan blades whose moment of inertia is I:$$\tau=I\alpha$$Given how long it takes for the fan blades to stop the frictional torques must be fairly low and so the torque applied by the motor to keep them going must also be low. With the relatively small torque rating, even if the ... 18 Calculating the power emitted as gravitational waves is relatively straightforward, and you'll find it described in any advanced work on GR. I found a nice description in Gravitational Waves: Sources, Detectors and Searches. To summarise an awful lot of algebra, the power emitted as gravitational waves by a rotating object is approximately:$$ P = ...

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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 ...

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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 ...

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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 ...

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A much simpler way of thinking about this is to consider energy. When the fan is spinning it has quite a lot of kinetic energy (try to stop it by putting your finger in the way to confirm this (don't actually do this!)). That kinetic energy goes as the square of the rotation rate, in fact. So as the fan starts, the motor needs to add energy to it. It ...

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You might be thinking in comparison to a desk or handheld electric fan. As mentioned by @Farcher, $\tau = I\alpha$. $I$, the moment of inertia of a spinning body around a particular axis of rotation, is calculated as follows: $$I = \iiint\rho(x,y,z)||r||^2\ dV$$ Or with uniform density, $$I = \rho\iiint||r||^2\ dV$$ From this formula, you can see that ...

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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. ...

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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 ...

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It's a bit complicated (Wikipedia). Induction motors work in sync with the AC frequency but have no torque at 0 RPM so they need some arrangement to get them started.

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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 The reason why a gyroscope does behave in this strange way is that if you try to rotate it's axis in some direction, the "endpoints" of this axis have to be pushed perpendicular to what our first intuition would say. In order to verify why the axis starts rotating in this strange way, let's make some simplifications: the gyroscope consists of two identical ... 6 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 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 ... 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 ... 5 The tangential acceleration a_t and the angular acceleration \dot{\omega} are basically the same thing. They are related by:$$ a_t = r\dot{\omega} $$So we don't include both of them because that would be counting the same thing twice. 5 In your energy conservation equation, your are assuming that both the initial system and the final system have kinetic energy due just to the rotation around the axis. However, there is also some kinetic energy due to the rings translating away from the axis. In other words, the velocity vector of a ring is not parallel to the velocity vector of the point ... 5 Because your pen is not a cylinder, but a portion of a cone. Since it is also rigid, both ends have to complete one cycle of rolling simultaneously. This means that for each cycle, if the narrow and thick ends are separated by the pen's length L and have radii r_1 and r_2, respectively, they roll 2\pi r_1 and 2\pi r_2, respectively. The only way ... 5 There are several factors that may be taken in account, but the more important is the energy used deforming the tire. Suppose a deflated tire. As you move forward and the tire rotates, the part of the tire that is starting to touch the ground has to be deformed (since the tire is flat). You have to use an important amount of energy for that. Note that the ... 5 simply the resistance of a body to rotate it over an axis? Gosh, I dislike the word resistance in this context since resistance is, in general, dissipative and, in particular, resistance to rotation would imply that an isolated object that is rotating would eventually stop. Think of moment of inertia (rotational inertia) about an axis as a measure of ... 4 Suppose you pedal at about 100 rpm (I don't know what a typical rate of pedalling is, but this seems a plausible order of magnitude). To make a fair comparison with your motorbike you need to gear the moorbike engine down from 7,000 rpm to the same 100 rpm that you pedal at, i.e. a factor of 70, and this will multiply the torque by a factor of 70. So at 100 ... 4 If the bearings were to be considered frictionless, then the maximum speed of the fan will not decrease, though it will take the fan longer to reach the maximum speed. Because as the moment of inertia of the impeller increases its angular acc. will decrease (for the same torque applied), therefore it will take the fan longer to reach its maximum speed. The ... 4 Do I need to use the angular velocity vector in the rotating or inertial reference frame for this? Yes. You can do it either way. I start with the expression that relates the time derivative of a vector quantity \boldsymbol u in the inertial and rotating frames:$$\left(\frac {d\boldsymbol u}{dt}\right)_I = \left(\frac {d\boldsymbol u}{dt}\right)_R ...

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