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How do you formally define angular velocity? Imagine a rigid body whose center of mass is fixed, and it is free to rotate about it. What are all the allowed motions? By definition, a rigid body has all the distances between separate particles fixed. Lemma 1 Image two arbitrary particles on a rigid body with locations $\boldsymbol{r}_i$ and $\boldsymbol{r}_j$...


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Both. Suppose that a particle of mass $m$ is in uniform circular motion with radius $r$ and tangential velocity $v_T$. We know that there must be a centripetal force maintaining this motion $$F_c = m \frac{v_T^2}{r}.$$ We also know that the system has an associated angular momentum whose magnitude is $$L = rmv_T.$$ Now if we only increase the centripetal ...


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For a point, you do talk about its velocity. By abuse of language, for an axis aligned with $\vec{n}$, I've seen people refer to angular velocity of the point about the axis as $\dot{\theta}\hat{n}$, that is, if you express the coordinates of the point in a cylindrical coordinate system with coordinates $(\rho, \theta, z)$ in which $\hat{n}$ is aligned with ...


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For a rigid object, the equation of motion for the center of mass $\mathbf x_\text{cm}$ of the object is described by Newton's second law $$\mathbf F_\text{net}=m\ddot{\mathbf x}_\text{cm}$$ Since your new composite system (ring + particle) has no forces acting on it after the collision, we have $0=m\ddot{\mathbf x}_\text{cm}$, i.e. the center of mass cannot ...


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How may I calculate earth's angular speed at specific point (The green dot for example)? The angular velocity doesn't depend on the position of the green dot. Assuming uniform rotational speed it is simply: $$\omega = \frac{2\pi}{T}$$ where $T$ is the period of rotation, in the case of Earth $T=24\mathrm{h}$. The tangential velocity $v$ for a point is given ...


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As long as there is a net velocity on the pendulum bob at the moment the elevator goes into free fall, the pendulum will go into uniform circular motion. The formula you have stated for the time period is only valid for a pendulum. Once the bob goes into circular motion, it is no longer a pendulum as there is no restoring force acting on the bob. The formula ...


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I think that one has to assume that there is no torque on the ball about the point it's rotating. So, it is like a pirouette problem \begin{align*} \mathbf{\tau}=\frac{d\mathbf{L}}{dt}=0&\Rightarrow \mathbf{L}\text{ is constant.}\\ \Rightarrow mv_1r_1&=mv_2r_2 \end{align*}


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Look at the forces acting on bob, when elevator's acceleration is $g$, in the axis which is perpendicular to velocity of pendulum. Let angle between rope and $y$ axis be $\theta$. So: $$\frac{mv^2}{l}=T+ma\cos(\theta)-mg\cos(\theta)\mathrel{\stackrel{{\mbox{ a=g}}}{=}}T$$ So: $\frac{mv^2}{l}=T$. And there's no force in direction of velocity there's only ...


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Notice, $V=R\omega$ is the linear or tangential velocity of any particle of a body rotating at angular speed $\omega$ at a particular instant of time $t$ because the tangential velocity $V=R\omega$ keeps on changing in direction from instant to instant while the magnitude $R\omega$ is constant if angular velocity $\omega$ of rotating body is constant. ...


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From Euler's theorem we know that a rigid motion in odd number of dimensions with one point fixed is a pure rotation about an axis (cf. Goldstein). This allows one to write, $$\mathbf{v}(\mathbf{r}) = \mathbf{v}_A + \omega_A \times(\mathbf{r}-\mathbf{r}_A), \quad \quad (1) $$ $$\mathbf{v}(\mathbf{r}) = \mathbf{v}_B + \omega_B \times(\mathbf{r}-\mathbf{r}_B). ...


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If $\omega$ be the angular velocity, then we can also write (besides the equation you have written for $\theta$), $$\omega=\omega_0+\alpha t\tag{1}$$ $\omega_0$ is the initial angular velocity of the circle at time $t=0$ . The angular acceleration is different from tangential or radial acceleration, i.e., linear acceleration $a$, in the same lines as angular ...


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