# Tag Info

7

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 ...

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 ...

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 ...

5

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 ...

5

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 ...

4

For the person not to slip, there must be a centripetal force of $mv^2/r = m r \omega^2$ towards the centre. Since $v$ varies with $r$ while $\omega$ is fixed ($v=r\omega$), it is probably easier to take the second form, in which case this force has to increase as $r$ increases. This forces comes from friction since there are no other forces in the plane ...

4

The total linear momentum of a system of particle labeled with $i \in {1\dots n}$ can be found in the microscopic view just by summing the linear momentum of the constituents: $$\vec{P} = \sum \vec{p} = \sum m_i \vec{v}_i$$ Now, writing $M = \sum m_i$ for the total mass, $\vec{X}$ as the position of the center of mass, and $V$ as the velocity of 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

Angular rotation is a vector so at any given instant any rigid body can only be rotating about one axis. If the body is rotating freely in space with no external forces then angular momentum is conserved. If the object is spherically symmetrical like the ball you suggest as an example, then the angular velocity is in the same direction as the angular ...

3

It looks just like rotation around a different axis with a different rotational speed. Specifically, if you set an object to rotate with angular velocity $\vec\omega_1$ and also with angular velocity $\vec\omega_2$, then it's really rotating with angular velocity $\vec\omega_1 + \vec\omega_2$. The direction of the vector $\vec\omega_1 + \vec\omega_2$ is the ...

3

As Manishearth says, for engines with more than one cylinder the firing of the other cylinders rotates the crankshaft. However, as any fan of vintage motorcycles will know, you can have four stroke engines with a single cylinder. In this case the engine has a heavy flywheel attached to the crankshaft and the momentum of the flywheel keeps the crankshaft ...

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

The definition of rotational kinetic energy is $$E_\text{rot}{}_{(i)} = \frac{1}{2} J_i \omega_i^2 = \frac{\;\; L_i^2}{2J_i}$$ where $J_i$ is moment of inertia, $\omega_i$ is angular velocity and $L_i = J_i\omega_i$ is angular momentum of the particle. If you select $\vec{r}_\text{cm}$ as the center of rotation these values can be calculated for each ...

2

You are right; the magnitudes of the translational and rotational kinetic energies are the same. In general, the kinetic energy of a mass distribution can be written $$K = K_{cm} + K_{rel}$$ $K$ is the total kinetic energy. $K_{cm}$ is the kinetic energy of a point particle whose mass is equal to the total mass of the distribution, if the point ...

2

A two-wheeled robot cannot move in the direction parallel to the axis of its wheels. Therefore, it cannot move in an arbitrary direction independent of its rotation; it is not holonomic. This implies that your robot cannot move in a straight line while spinning about its center (as a weightless/frictionless body would). However, it sounds like you are OK ...

2

This is actually fairly straightforward, at least based on what you've described. You have a left wheel which moves at an angular velocity $\omega_L$ and a right wheel which moves at an angular velocity $\omega_R$. The distance traveled by the edge of each wheel in a time $t$ is $r\omega t$, where $r$ is the radius of the wheel, and accordingly $r\omega t$ ...

2

The problem is exactly analogous to a very famous problem which is presnt in any good Rotational book, Whats happening here is ball has a velocity of its centre of mass,Vcm, and an angular velocity. Now when we put the ball on a friction ON surface, the force of friction provides a torque and an acceleration to the centre of mass, this happens till the ...

2

In an internal combustion engine, we have multiple cylinders. They are attached to a shaft in an alternating manner such that when one set of the cylinders have combustion, they drive the shaft to move down in the other set. See http://commons.wikimedia.org/wiki/File:Cshaft.gif

2

This expression means that when considering rotational motion one cannot make the simplification of assuming that the rotating body in consideration is a point mass. In other words: shape and mass distribution matters. This is due to the fact that spatial distribution of mass of the body around its rotation axis is a very important factor determining the ...

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

Similar to the derivation of separation of angular momentum into $L_{CM}$ and $L_{internal}$, one can derive similar expression for Energy as $E = \frac{1}{2}M_{total}v_{CM}^{2} + \frac{1}{2}\sum \mu_{i} v_{i}^{'2}$. Proof: $$E = \frac{1}{2}\sum \mu_{i} v_{i}^{2}$$ $$v_{i} = v_{CM} + v_{i}^{'}$$ $$E = \frac{1}{2}\sum \mu_{i} v_{CM}^{2} + v_{CM}\sum ... 1 From Goldstein, chapter 4 eqn 4-92', for a finite rotation the change \boldsymbol{\Delta r} caused by rotating a vector \boldsymbol{r} through an angle \Phi about a direction defined by a unit vector \boldsymbol{n} (\Phi positive for a counter-clockwise rotation), to a final position \boldsymbol{r'} is given by:$$ \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". ...

1

You can prove it with a little trick. For a particle on a circle of radius $r$ we have $\vec{x}(t)\cdot\vec{x}(t)=r^2$ at each point in time. Differentiating with respect to time we get $\vec{x}\cdot\vec{v}=0$. Differentiating again we get $v^2+\vec{x}\cdot\vec{a}=0$. But $\vec{x}\cdot\vec{a} = - r a_{\mathrm{rad}}$, and the relation \$a_{\mathrm{rad}} = ...

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