# Tag Info

19

Let me first list all of the possibilities I considered that I later rejected. This is far from exhaustive, and I'm looking forward to seeing other people's creativity. Bad Ideas Sit on a tire swing with the fan pointing to the side. Point the fan up, measure speed of rotation of the system on the tire swing. Get a laser or collimated flashlight. Point ...

9

There are no "other" examples. The condition that $\vec \omega$ and $$\vec L = I_{\rm tensor} \cdot \vec \omega$$ point to the same direction i.e. $$(\vec L=) I_{\rm tensor} \cdot \vec \omega = k \vec \omega$$ where $k$ is a real number (and no longer a tensor) is a definition of an eigenvector of $I_{\rm tensor}$: both $\vec \omega$ and $\vec L$ are ...

9

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

There is indeed a term involving the time derivative of the changing coupling between the masses. First, let's derive the equation for a single mass. $$L = \frac{1}{2} I\, \dot{\theta}^2 - V(\theta)$$ $$\frac{\partial L}{\partial \dot{\theta}} = I\, \dot{\theta}$$ $$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$ $$\tau = \frac{d}{dt} ... 7 There are actually several different ways to interpret that question, depending on what you mean by "vector" and "rotation". But here's a sense that I've often wondered about myself: in introductory physics, the velocity vector is defined as the time derivative of the position vector (relative to some fixed point). Why is the same not true of angular ... 6 You made a mistake in assuming that the angular acceleration (\alpha) is equal to v^2/r which actually is the centripetal acceleration. In simple words, angular acceleration is the rate of change of angular velocity, which further is the rate of change of the angle \theta. This is very similar to how the linear acceleration is defined. ... 6 Defining properties of vectors are that you can add them and multiply them by constants. These both make sense for angular velocities. On the other hand, adding rotations doesn't make sense. What you can do with two rotations is compose them: first rotate one way, then rotate another. This operation doesn't look like addition of any sort. For one thing, it ... 6 Here's a straightforward but somewhat computational way. There are two steps. (1) Show how to define the angular velocity vector in terms of rotation matrices. (2) Write a general rotation in terms of Euler angles. (3) Combine (1) and (2) to get an expression for the angular velocity vector in terms of Euler angles. Step 1. Recall that if \mathbf x(t) is ... 5 Yes there will be a drag torque opposite the direction of spin. The name for this seems to be viscous torque. See e.g. this paper. Now according to this paper, the viscous torque on a spinning sphere of radius R in a fluid with viscosity \eta spinning with constant angular velocity \vec\Omega is$$ \vec\tau = -8\pi R^3\eta\vec\Omega $$The paper ... 5 two concentric and counterrotating flywheels preclude all precession forces regardless of which plane the axis is rotated in. this is assuming the connection between the two flywheels is sufficiently strong--it make break from tension/compression due to each flywheel experiencing its own forces. refer to the diagram i just drew up. the black rectangles ... 5 I will attempt to answer this question with some basic dynamics and some contact mechanics. There are two special cases here. a) There is sufficient friction to keep the base of the pin A fixed (imparting a reaction impulse J_A when hit by the ball, or b) The floor is smooth and the pin will translate and rotate at the same time with J_A=0. There is ... 4 For the person not to slip, there must be a centripetal force of mv^2/r = m r \omega^2 towards the centre. Since v varies with r while \omega is fixed (v=r\omega), it is probably easier to take the second form, in which case this force has to increase as r increases. This forces comes from friction since there are no other forces in the plane ... 4 a_c = \frac{v^2}{r} isn't angular acceleration. It's the magnitude of the linear acceleration towards the centre of an object following a circular path at constant angular velocity. Angular acceleration is the derivative of angular velocity, and the analogue of Newton's second law is that angular acceleration equals torque divided by moment of inertia. 4 Assuming your rotating object (e.g. the Earth) is rotating at a steady speed the only way to change it's apparent speed of rotation is if you're rotating around it. You give the example of a geostationary satellite. This rotates around the Earth at the same angular velocity as the Earth rotates, so the Earth appears to be stationary (hence the name ... 4 I assume you know about rotation matrices, and so for a sequence rotations about Z-X-Z with angles \phi, \theta and \psi repsectively you have$$ \vec{\omega} = \dot{\phi} \hat{z} + T_1 \left( \dot{\theta} \hat{x} + T_2 \left( \dot{\psi} \hat{z} \right) \right) $$The logic here is apply a local spin of \dot{\phi}, \dot{\theta} and \dot{\psi} ... 4 The moment of inertia tensor is not constant in the external reference frame (http://en.wikipedia.org/wiki/Precession#Torque-free ) 4 High angular momentum presents a barrier preventing collapse to a black hole (at least until this angular momentum is radiated away). The parameter on which the formation of black hole depends is the ratio q of angular momentum (J) to the square of mass (M). If q=J/M^2 < 1 (in relativistic units with G=1, c=1), then the black hole ... 4 The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors \hat{i} and \hat{j} as well as the rotating polar unit vectors \hat{e}_r and \hat{e}_\theta. The Cartesian unit vectors \hat{i} and \hat{j} are stationary and ... 3 There are lots of different examples of oscillatory systems that have essentially the same mathematical form. Let's start by just looking at one type of differential equation: a = \frac{d^2 x}{dt^2} = -\omega^2 x This equation has a general solution (you can check this) x(t) = A \sin (\omega t + \phi) which oscillates with a period of ... 3 The g force is a unit of acceleration. 1 g is equal to 9.80665 m s-2. So the correct formula is$$ \text{G force} = \frac{\text{Acceleration in m s}^{-2}}{9.8}. $$However, when describing uniform circular motion (i.e. \boldsymbol\omega is constant) in free space, the only acceleration felt by the person rotating (in their frame of reference) is the ... 3 The condition for staying in a circular orbit is the requirement for the centripetal force to be equal in magnitude to the gravitational pull. To be precise:$$F_g=F_c,mg=\frac{mv^2}{r},$$where F_g is the absolute value of gravitational force, F_c the absolute value of centripetal force, g the gravitational acceleration, m the mass of the ... 3 In the basic discussion of angular momentum where something is rotating around a fixed symmetrical axis \vec{L}=\vec{r}\times\vec{p} reduces to \vec{L}=I*\vec{\omega} Like in this animation where each vector is colored appropriately: However angular velocity and angular momentum can have different directions in two cases: If the axis of ... 3 I'm outlining this and stating the final result so that the OP gets the fun of figuring this out themselves. Future responders, please don't work this out All you have to do is allow \omega(t) to be a function of time. You'll get extra {\dot \omega} = \alpha terms in your equation, and you'll get a final result that says that$${\vec a} = {\vec ...

2

Always start with the units. They'll tell you a lot about the equations, and allow you to fix consistency errors. Incidentally, this is why I prefer Leibniz's notation over Newton's for derivatives, the units are immediately determined by examining the derivative, e.g. $dx/dt$ has units of distance over time assuming the usual definition of $x$ and $t$. ...

2

A Newton is 1 ${\rm kg\cdot \rm m/s^2}$, not 1 ${\rm kg\cdot s^2/m}$. Also, you need to take the square root at the end.

2

In general, the angular momentum for a rigid body is $\vec{L}=I\vec{\omega}$. For the special case of a point particle $\vec{r}$ from the axis of rotation, we have $I=mr^2$ and $\vec{\omega}=\frac{\hat{r}\times\vec{v}}{r}$, or $\omega=\frac{v}{r}\sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{r}$. In this case, the angular momentum ...

2

They do not come closer to each other according to the picture. They always keep the distance of the $3L$ since they hold onto that bar that is going to rotate counter clockwise. I think your answer $$\omega = \frac{2}{3} \frac{v}{L}$$ is fine. This works since $v$ and $r$ are perpendicular, with a 90° angle in between.

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible