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65

The Foucault pendulum is a great experiment which does demonstrate that the Earth is rotating, but it was only introduced in 1851. The Earth had been known to rotate for several centuries before that, probably stimulated by Copernicus and Galileo pushing the heliocentric model of the solar system during the 16th century. A couple of decades before Faucalt's ...

49

Foucault pendulum. I don't know how the ancients did it, but it is surely pure classical mechanics. The animation describes the motion of a Foucault Pendulum at a latitude of 30°N.

46

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 ...

20

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 ...

16

I think the Foucault pendulum is the best answer, but for the sake of variety I'll add another very interesting one: the equatorial bulge affecting the figure of the Earth. This is the "pancaking" of the planet due to its rotation. You can measure the geometry of the Earth without leaving its surface, and find that it is bulging in accord with your ...

13

From conservation of angular momentum we have $(I+\Delta I)(\omega+\Delta \omega) = I\omega,$ or $$\frac{\Delta \omega}{\omega} = - \frac{\Delta I}{I+\Delta I} \simeq -\frac{\Delta I}{I}.$$ We make the following simplifying assumptions: The earth is a sphere of uniform density of mass $M$ and radius $R$. The building is constructed on the equator by ...

12

An indirect indication that the Earth rotates is the fact that the rotation varies over time. First of all, the orientation of the Earth's axis changes: long-term effects like precession and slow variations in the axial tilt, as well as small short-term variations like nutation. Precession was already known in the Ancient world (Hipparchus, Ptolemy,...) and ...

11

The system needs to conserve momentum. In both cases, the momentum is whatever m*v is for the bullet. Since it's the same in both cases, the bullet and block have the same vertical velocity. Mechanical energy is not conserved. The reason the block hit on the side has more kinetic energy is that the bullet converted less of its kinetic energy into heat upon ...

10

Well, if we make a quick estimate of the mass of a huge building. Let's say the building has a base of $100\times100 \;\text{m}^2$ and a height of $1500 \;\text{m}$, this is already substantially bigger than the current biggest building. Then we have a volume of $1.5\times 10^7\text{m}^3$. If we make the assumption, again very rough and on the high side, ...

8

ZPM isn't the full answer. A combination of gyro and thrusters are used. Primarily they use Control Moment Gyroscopes (CMG) located in the Unity Module. Secondary options with more thrust are the Russian Control Thrusters on both Progress and Zvezda (means star) modules. The CMG's are quite heavy gyroscopes at about 600 lbs each. Inside the black ...

7

The moment-of-inertia (MOI) tensor is real (no imaginary terms), symmetric, and positive-definite. Linear algebra tells us that for any (3x3) matrix that has those three properties, there's always a set of three perpendicular axes such that the MOI tensor can be expressed as a diagonal tensor in the basis of those axes. These are called the principal axes ...

7

The rectangular prism is a rigid body. The equations of motion of a rigid body around its center of mass are given by: (Please, see for example: Marsden and Ratiu , (page 6). $$I_1\dot\Omega_1=(I_2-I_3)\Omega_2\Omega_3$$ $$I_2\dot\Omega_2=(I_3-I_1)\Omega_3\Omega_1$$ $$I_3\dot\Omega_3=(I_1-I_2)\Omega_1\Omega_2$$ Where $\Omega_1,_2,_3$ are the angular ...

7

Here are some supporting evidence, taken from here. The inner core rotates in the same direction as the Earth and slightly faster, completing its once-a-day rotation about two-thirds of a second faster than the entire Earth. Over the past 100 years that extra speed has gained the core a quarter-turn on the planet as a whole, the scientists found. Such ...

7

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 ...

6

The instability inherent in the medium length axis or $\prod_2$ as shown above is discussed in detail in Marsden and Ratiu, which is where the image is from. The unstable homoclinic orbit that connect the two unstable points have intersting features. Not only are they interesting because of the chaotic solutions via the Poincare-Melnikov method that ...

5

You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling. What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily ...

5

As in the answer of @Mark Eichenlaub the mass of the lawnmower won't increase! Of course the blades of the lawnmower can have a pull-effect in which they might aid your foreward-movement (in which degree this might help, I'm not sure ...). Of course a rotating blade creates an angular momentum, so if you were to make a turn with your lawnmower you'll need ...

5

If you start in the rest frame of the wheel the velocities of the top and bottom points are $v$ and $-v$, and the velocity of the centre of mass (black dot) is of course zero because that's how we define the rest frame. If this wheel is on a moving vehicle the velocity of the bottom must be zero, because it's in contact with the stationary road. To make ...

5

I have observed this as well, and experiment suggests it's because the dust is hydrophobic. If you splash a small amount of water gently onto the dusty surface you will see the water roll up to form beads that do not wet the surface. This is my rather crude attempt to illustrate what happens when you try and wet the dusty surface: The brown splodges are ...

4

The electric field is nonzero. For a cylinder of finite length, it's nonvanishing everywhere. In the limiting case of an infinitely long cylinder, the field is only nonvanishing inside the cylinder. The easiest way to solve this is to use the fact that the electric and magnetic polarizations $(-\textbf{P},\textbf{M})$ transform in exactly the same way as ...

4

The angular momentum of each disk individually is not conserved, however the total angular momentum of both disks is conserved because there are no external torques acting. Start by calculating the total angular momentum of both disks (I'm going to replace "w" by "v" since "w" is confusingly close to "$\omega$"): \begin{align} L_{total} &= I_a ...

4

Measuring the geometry of the earth, we find that it has an equatorial bulge. We make no assumptions about the cause of the bulge, though it suggests already that the earth is rotating as @Mike has described. We measure the acceleration due to gravity at the poles and on the equator. Most of the difference we find is accounted for by the bulge, but there ...

4

The direction of angular velocity is different from that of regular velocity for (arguably) two reasons. First, it points out of the plane because of the nature of angular velocity. It signifies a rotation, as such, there is not any particular direction unit vector in every coordinate space that could represent it. In spherical or cylindrical coordinates, it ...

4

Hint: Look at the following diagram, and then solve the equations: Or just notice that $F_w$ and $F_f$ do not depend on $\omega$, then use the Work-Energy principle. step by step solution:

4

The components of any vector function can be written any any desired basis. In particular, let \begin{align} \mathbf A_L(t) = (A^1_L(t) , A^2_L(t), A^3_L(t)) \end{align} denote the components of a vector function as written in an orthonormal basis fixed in the laboratory, and let \begin{align} \mathbf A_R(t) = (A^1_R(t), A^2_R(t), A^3_R(t)) \end{align} ...

4

No, a lawn mower is not heavier when spinning in any significant way. If it is harder to push, this is probably because of friction working against rotation of the wheels. In theory, the lawn mower has slightly-higher mass when spinning according to $E = mc^2$. For a spinning lawn mower this is on the order of $10^{-16}$ the mower's weight, or $10^{-12}$ ...

3

OK, the direction of precession of a gyroscope. The first image shows a gimbal mounted gyroscope wheel. From outside to inside there is a yellow housing and a red housing. I define three axes: Roll axis - the gyroscope wheel spins around the roll axis. Pitch axis - motion of the red housing. As you can see, the gimbal mounting ensures the pitch axis ...

3

It does have a force. It is the centripetal force. Usually, the spinning body is a circular object, so all centripetal forces are balanced out to 0. For each point mass, there is another point mass on the opposite side. It is the symmetry of the system that keeps everything in the same place spinning.

3

You have already answered your own question! There is a force between the hinge and the door. If the door weren't attached to the hinge, it would start flying away in addition to spinning. The only error you have is a mistake in your integral. The net force on the door is $\frac{aML}{2}$ in your notation. (Note: it is probably best not to use $a$ for both ...

3

Let us check the claim that the earth is rotating about it's own axis. We may choose this axis to be the $z$-axis. The earth can be approximated by a sphere. Consider a pendulum living somewhere on the surface of the earth, initially swinging on a north-south line. The position of the bob is described a vector ${V} = (V^{\theta}, V^{\phi})$ living in the ...

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