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56

A hollow sphere will have a much larger moment of inertia than a uniform sphere of the same size and the same mass. If this seems counterintuitive, you probably carry a mental image of creating the hollow sphere by removing internal mass from the uniform sphere. This is an incorrect image, as such a process would create a hollow sphere of much lighter ...


35

Here's an illustration of a uniform sphere and a hollow sphere mid-sections with the same mass, if you better understand these things visually:


32

The key is... the closer the mass to the axis of rotation, the easier it is to add angular velocity to the body. For instance a figure skater rotates faster when she puts her limbs closer to her body. Let's see how it works in a more intuitive fashion: For instance, in the figure bellow, trying to lift up table (A) would be easier than table (B). In ...


25

It depends what you mean by day and night. The day and night are not of equal lengths now, where I live at latitude 53N. The tilt of the Earth's rotation axis with respect to the ecliptic plane means that this is generally true. The situation you describe would have to be considerably more extreme. If the planet was in a highly eccentric orbit and had a ...


24

I know that the opposite could happen. There is an old book called "Night Fall" about a planet that had three stars. Because there was always a star shinning on all sides it would never get dark. About every 500 years everything lines up just right so that all the stars were on one side. Then as the planet rotated on its normal spin, night fall would come ...


23

"A state of rest" is a relative term. Relative means - measured in comparison to the things around it. When you sit in a train and sip from a cup of coffee, you can do so because the cup is still relative to you even though both of you might be hurtling through the countryside at 200 km/h. For most experiments, objects can be considered "at rest" if they ...


22

Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right. Shift the ...


18

The moment of inertia of a body about an axis is a measure of how far the mass is distributed from that point. For a solid sphere of mass $m$, radius $r$, you have the mass distributed continuously from the center to the radius. However, for a hollow sphere of mass $m$, inner radius $r_i$ and outer radius the same as before, $r$, you have all the mass ...


15

In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, ...


14

In principle, yes, but the effects are almost completely negligible. As objects on the surface of the earth move around, the Earth's moment of inertia changes by minute amounts, and this affects its rotation. However, performing an order of magnitude estimate on the ratio of the contribution to the moment of inertia of a person-sized object on the equator ...


12

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 ...


11

The equation you are referring to is the expression for the moment of inertia of a point particle of mass $m$ at a distance $R$ away from some axis. This is expression is really the definition of the moment of inertia for a point mass, so the question becomes "where does this definition come from, and why is it useful?" Well for simplicity, suppose that ...


11

This is a note on why angular velocities are vectors, to complement Matt and David's excellent explanations of why rotations are not. When we say something has a certain angular velocity $\vec{\omega_1}$, we mean that each part of the thing has a position-dependent velocity $\vec{v_1}(\vec{r}) = \vec{\omega_1} \times \vec{r}$. We might consider another ...


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 ...


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 ...


9

One may prove for an arbitrary rigid body (and wrt. to an arbitrary choice of pivotal point for the rigid body) that the three moments of inertia $I_x$, $I_y$, and $I_z$, around the three principal axes (which we will call $x$, $y$, and $z$) satisfy the triangle inequality, $$\tag{1} I_x +I_y ~\geq~ I_z, \qquad I_y +I_z ~\geq~ I_x, \qquad I_z +I_x ~\geq~ ...


8

I'll tackle your questions in reverse: 3. The contact point is stationary because the wheel is not slipping. This happens when the force of static friction is able to counter the force of the wheel on the ground. This is what you want for controllable transport. If the wheel starts slipping (because of low friction) that's a skid and you are no longer able ...


8

The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and ...


8

Radian $(\theta)$ is defined as, $\theta=\dfrac{l}{r}$, where $l$ is length of arc and $r$ is radius in a circle, and both have dimension as lengths. Thus, Radian is a dimensionless unit.


7

If planet is going around some star, at least somewhere at the surface of this planet is going to be a day, or else some other celestial body is covering the star light. If the speed of rotation of the planet is arranged in such a way that while it's traveling around the star one side of planet is always facing the star, and other side is always facing the ...


7

There is a identity for the derivative of the cross-product of two vector functions $\mathbf A(t)$ and $\mathbf B(t)$; \begin{align} \frac{d}{dt} (\mathbf A \times \mathbf B) = \frac{d\mathbf A}{dt}\times \mathbf B + \mathbf A\times \frac{d\mathbf B}{dt} \end{align} Using this rule with the computation you're considering, we obtain \begin{align} ...


7

In the basic discussion of angular momentum where something is rotating around a fixed symmetrical axis $\vec{L}=\vec{r}\times\vec{p}$ reduces to $\vec{L}=I*\vec{\omega}$ Like in this animation where each vector is colored appropriately: However angular velocity and angular momentum can have different directions in two cases: If the axis of ...


7

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 ...


7

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun. However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, ...


7

There are actually several different ways to interpret that question, depending on what you mean by "vector" and "rotation". But here's a sense that I've often wondered about myself: in introductory physics, the velocity vector is defined as the time derivative of the position vector (relative to some fixed point). Why is the same not true of angular ...


7

I agree with the previous answer. Angular momentum, something the earth has because of its rotation about its axis, can only be changed when an external torque (twisting motion) is applied to the earth. As far as I know, there are two ways in which this can happen. If there was friction between the earth's surface and space, then that would slow down the ...


7

There is no gravitational waves for a uniformly rotating axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fields, §88 The constant gravitational field: However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the ...


6

One way to explain it that makes sense to me is that the backwards spin on the ball is a force pushing the ball toward you. The one time applied force of your finger that creates the backward spin also pushes the ball forward, but the backspin stays almost constant. At the moment you spin the ball, the forward force is greater then that of the backspin, so ...


6

Defining properties of vectors are that you can add them and multiply them by constants. These both make sense for angular velocities. On the other hand, adding rotations doesn't make sense. What you can do with two rotations is compose them: first rotate one way, then rotate another. This operation doesn't look like addition of any sort. For one thing, it ...


6

The inertia ellipsoid is computed from an integral about an axis - in other words you rotate the object. This will "smooth out" any symmetries and typically increase the symmetry. Sorry this is a "early morning" intuitive explanation - maybe someone else will give you a more formal answer.



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