# Tag Info

7

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 ...

6

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 ...

4

In principle, yes, but the effects are almost certainly 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 ...

3

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, ...

2

Let $\vec r_0(t)$ denote the point around which the object is rotating and $\vec r(t)$ the position of the object. Then the fact that the particle is rotating around the point $\vec r_0(t)$ can be formalized by the mathematical statement that $$\vec r(t) - \vec r_0(t) = R(t) \vec c$$ for some constant vector $\vec c$ and time-dependent rotation $R(t)$. ...

2

Your angular velocity vector is $$\vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|}$$ where $\vec{r}_A = (0,0.2,0.12)$, $\vec{r}_D = (0.3,0,0)$, $\vec{r}_B = (0.3,0.2,0.12)$ in meters and $\Omega = 90\;{\rm rad/s}$. Your velocity kinematics is $$\vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A )$$ And acceleration ...

1

Because you've not employed the right angles yet. The velocity of something moving in a circle is necessarily tangent to the circle at whatever point it lies. That property means that the velocity vector is at a right angle to the radius vector. The radii vectors are AB and AC, and their corresponding velocity vectors are v1 and v2. We know that the ...

1

It does, but changes are so miniscule that they don't really count. Even if everyone in the world started running in the same direction (East, for example), their total angular momentum would be on the order of $10^{18} \ \text{N m s}$, in comparison to the earth, which is on the order of $10^{33} \ \text{N m s}$. We would change the earth's rotation by ...

1

The relation between angular velocity $\vec{\omega}$, position $\vec{r}$ (assuming rotation around the origin) and tangential velocity $\vec{v}$ (which is what you are asking for) is given by $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{\mid\vec{r}\mid^2},$ where $\times$ is the cross product and $\mid\vec{r}\mid^2$ the norm of the position vector squared. ...

1

The outcomes are not supposed to be the same. There are two ways to interpret your question: 1. You want to calculate the kinetic energy in different reference frames. Let's think for example of a point-like body moving in a constant velocity $\mathbf{v}$. It's kinetic energy is $\frac{1}{2}mv^2$, but if we calculate it in a reference frame that is moving ...

1

It's simple: the body with negative angular acceleration has a decreasing angular velocity (or an increasing negative angular velocity). Mathematically, it's simple: if $\omega$ is the angular velocity (and $\omega_0$ is the angular velocity at $t=0$, then it follows the relation $$\omega=\omega_0+\alpha\times t$$. If $\alpha$ (angular acceleration) is ...

1

I'll compare the equations of motion of a point particle and a rigid body. To understand a free point particle, once I solve its motion in one dimension ($x(t) = v_0 t + x_0$) I can immediately solve the three dimensional problem. This is because the equations are separable and commutative: since translations along $x$ (label this symmetry $T_x$) implies a ...

1

My answer: 816.126 nanometers. Someone please review my thought process before I accept my own answer + declare myself awesome. OK, I originally asked this question to save myself from doing some math, but this was clearly a bad idea, as it doubtless angered the Math Gods. I've now come up with an answer, which, of course, agrees with neither of the ...

1

Rotating bodies emit gravitational waves. The emission is of quadrupole radiation. The radiation power (which is proportional to the fifth power of the angular speed) causes gradual reduction of the angular speed. Please see the following review by: Alessandra Buonanno equations 7.14, 7.17.

1

The formulae from the NASA document and Wikipedia are simply in different frames: the ones in Wikipedia assume coordinate frame whose axes pass through the center of mass whereas the NASA document assumes arbitrary cartesian coordinate frame. Let capital Xi, Yi and Zi denote coordinates of i-th atom in an arbitrary Cartesian coordinate frame O and xi, yi ...

1

I fussed about this as well. My resolution: for these calculations the fixed-body frame is not to be considered as co-moving with the body, but rather a non-rotating frame that instantaneously aligns with the body. The Euler angles translate between the body and the space frames. The Euler angles are indeed functions of time, and the fixed-body frame is ...

1

This may not be intuitive at first but I think it is valuable in understanding the relationship between rotation matrices and angular velocities. Also, I know it does not direction answer the question, but I sense there is confusion in the OP and this might help. So given the rotation matrices $E_1$ and $E_2$ for two connected rigid bodies how do be ...

1

You are mixing up different things. The first is a rotation transformation. Such a transformation is linear and can therefore be written as a matrix. $$\bf\vec x'=A\vec x$$ Now, angular velocity, is the velocity of a physical rotation. $$\vec\omega=\frac{\mbox d\vec\theta}{\mbox dt}$$ This theta is the angular displacement, or the angle of rotation. This ...

1

Angular momentum is given by $L=Iw$ where $I$ is the moment of inertia. A turntable can be well approximated by a solid cylinder of radius $r$, height $h$ and mass $m$. Its moments of inertia are $$I_z = \frac{mr^2}{2}$$ $$I_x = I_y = \frac{m}{12} \left(3r^2+h^2\right)$$ Now you can see what happens when $m \rightarrow 2m$.

1

The moment of inertia is merely a generalisation/application of the ‘usual’ inertia to rotations. Since translations and rotations are different kinds of motion, it appears sensible (to me) to have different kinds of inertia associated with them. Regarding your second question: Imagine a particle at position $(x,0,0)$ which you would like to rotate with ...

1

Let's look at the given informations. We have an initial angular velocity, and we know how many revolutions it takes to get to zero angular velocity. So we have $\omega_i$, $\omega_f$, and $\theta$. Looking at that list of givens I would guess to use $$\omega_f = \alpha t+\omega_i \quad\text{and}\quad \theta_f=\frac{1}{2}\alpha t^2 + \omega_i t + ... 1 Say you're looking at the piece of paper on your desk; that is the xy plane. You place a dot in the center of the paper; that's your origin. Your angular momentum is \vec{L}=\vec{r}{\times}m\vec{v}. For this example, \vec{\omega} points in the same direction as your angular momentum, because \vec{L}=mr^2\vec{\omega}. The way I remember the ... 1 The least ambiguous way to describe it would be using cartesian vector notation. so we have \hat x, \hat y and \hat z. If the particle is moving in a circle from the \hat x axis to the \hat y axis, it's moving counter clockwise, like moving from 3 to 12 on an analog clock. We say it's angular velocity is in the + \hat z direction, or up out of ... 1 The rate of change of the rotation matrix \boldsymbol R is$$ \dot{\boldsymbol{R}} = \vec{\omega} \times \boldsymbol{R} $$where \vec\omega\times = \begin{pmatrix} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \end{pmatrix} is the cross product operator in 3x3 form. So if the position of a ... 1 From Goldstein, chapter 4 eqn 4-92', for a finite rotation the change \boldsymbol{\Delta r} caused by rotating a vector \boldsymbol{r} through an angle \Phi about a direction defined by a unit vector \boldsymbol{n} (\Phi positive for a counter-clockwise rotation), to a final position \boldsymbol{r'} is given by:$$ \boldsymbol{\Delta r} = ...

1

Always start with a nice clear diagram/sketch of the problem. It all follows from there. Here is a Free Body Diagram I made for you. Then you have (the long detailed way): Sum of the forces on body equals mass times acceleration at the center of gravity. $\sum_i \vec{F}_i = m \vec{a}_C$ $$A_x = m a_x \\ A_y - m g = m a_y$$ Sum of torques about ...

1

The accelerometers always tell you the 3 components of acceleration - including gravity. If the aircraft is on the ground, and the Z accelerometer is truly vertical, it will have an output equal to the value of gravity at that point on the earth. If you know the value of gravity at that point, and subtract that from the Z accelerometer output, the result ...

1

Using this picture as the standard for pitch, roll and yaw and let x be the forward direction, y be the side, and z be vertical, then geometrically, it looks like the true vertical acceleartion is $$a_{z}\left[\cos(Pitch)+\cos(Roll)\right]+a_{x}\sin(Pitch)+a_{y}\sin(Roll)$$ Didn't I just answer it for you yesterday at the math stack exchange? ...

1

Well, here's my epee: The problem is that basis vectors transform oppositely to a vector's components. Sticking to an active approach — wherein there is only one basis $\{\hat{e}_i\}$, and starting with the well-worn $r^\prime_i=R_{ij}r_j$, $$\vec{r}^\prime=R\vec{r} = R\left(\hat{e}_jr_j\right)= \hat{e}_i R_{ij} r_j ~,$$ we see that a column of basis ...

1

Seriously, by far the easiest way is to find it empirically. Take several coins and find the number of coins it is required to make a fair one by piling them together. The problem seems to be hard from the theoretical point of view. However, a bit of googling in google scholar brought out "Probability and dynamics in the toss of a non-bouncing thick coin". ...

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