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1

As MSha has explained, centrifuges won't work. The only way to protect astronauts from g-forces is to transfer the normal force that act on their bodies in a different way. The problem is that normal forces only act on the boundary surface. Obviously the entire volume of the astronauts will be accelerated, this means that that the force acting on each volume ...


2

If you place the astronauts in a centrifuge, then at some point you're actually increasing the G's that they experience. Think about it. If the centrifuge spins about a horizontal axis, then when the astronaut reaches the bottom of that spin, they'll feel extra force downward, even though they might be weightless at the top of the spin. If the centrifuge ...


0

If you have found that the initial angular acceleration of the disc is $\alpha = \frac {2g}{3R}$ that must mean that the initial centre of mass acceleration $a = \frac {2g}{3}$ because $a = R \alpha$. If the force acting on the pivot is $X$ upwards then applying Newton's second law for the centre of mass motion gives $mg-X = ma$, so $X = \frac {mg}{3}$. ...


0

As the puck is moving in uniform circular motion the centripetal force is normal to the velocity vector, no work is produced and the total energy of the puck, which is kinetic=$(1/2)mv^{2}$, remains constant. Pulling the string, even slowly, the string tension becomes oblique to the velocity vector during the transition, work is produced transfered to the ...


4

Without loss of generality, we assume that a particle is moving in a circle of radius $R$ centered at the origin and lying in the $x$-$y$ plane. Using cylindrical coordinates $(\rho, \phi, z)$, the angular velocity of the motion is $\boldsymbol\omega = \omega \hat{\mathbf z}$. The velocity is tangent to the circle and given by $\mathbf v = ...


2

In your scenario, angular momentum $m v r$ is preserved (because your pulling force is radial, with no tangential component). So if you reduce $r$ by half, $v$ must double, and since $\omega = v/r$, it increases by a factor of four. Note this means in a small amount of time that the area swept out by the string is proportional to $v$ and $r$. Since they ...


4

I think that what your teacher has told you is that the angular momentum of a body can be split into two components: The spin angular momentum which is an intrinsic property of the body and is independent of the point about which you wish to find the angular momentum. $L_{\text{spin}} = I_{\text{cm}} \omega = \frac v r $ where $I_{\text{cm}}$ is the ...


0

There is an exact solution to the problem. You can consider each point on the sphere by specifying two parameters ($r$ and $\theta$). Now since you know each points velocity vector you can calculate its contribution to the angular momentum by using $dL$ = $dm$($r$ x $v$). Now integrate over $\theta$(0 to $2\pi$) and $r$(0 to R). After doing all this, you ...


0

Are you confused about how you get into this? $$\begin{align} v & = \omega\,r \\ ({\rm m/s}) & = ({\rm rad/s})\,({\rm m}) = ({\rm m\,rad/s}) \end{align} $$ Radians are not units with dimensions. They can be seen as $({\rm rad}) = ({\rm m/m})$ like arc length to radius. This makes to above right hand side equivalent to the left hand side.


-1

This is all true because the ratio of a circle's diameter to circumference is constant (pi). The unit of radian is actually chosen so that l = r.theta. So if you go theta radians around a circle you travel r.theta. For the angular to linear velocities, think of a disc rotating at an angular speed omaga. then the further out from the centre of the disc you ...


1

You can look at this situation in two ways. First in terms of absolute motions $\Omega$ and $\omega$, and secondly it terms of the relative motion $\dot{q} = \omega - \Omega$. The results are the same The cylinder is spinning with $\omega$ speed and translating with $\Omega R$ speed. The combined momentum and kinetic energy is Linear Momentum: $$p = m R ...


0

Yes.. The two angular velocities are independent of each other... It's basically like earth rotating around the Sun... Earth's speed of revolution around the Sun is independent of its speed of rotation about it's own axis.... I'm not very sure about your first equation where you add the two angular velocities... As Leverl said, the two motions are ...


2

It has already been somewhat answered in this other question of yours: Parallel axis theorem and Koenig theorem for angular momentum (see also my comment to the answer). Angular velocity, like translational velocity, has to be defined by reference to a frame. In your question, you state the angular velocity about the principal axis is $ω$. So its angular ...


1

What you are referring to is a special case of the Ehrenfest Paradox In its original formulation as presented by Paul Ehrenfest 1909 in relation to the concept of Born rigidity within special relativity,1 it discusses an ideally rigid cylinder that is made to rotate about its axis of symmetry. The radius R as seen in the laboratory frame is ...


1

Well, my reasoning may be a bit circular, but I'd say that for the linear and angular velocity not to be independent, the one needs to be a function of the other: $v_{cm}=v_{cm}(\Omega)$. I can only think of cases where the relation would be directly proportional. Two examples: A ball or cylinder rolling over a surface without slipping: in this case ...


0

A particle following a prescribed path has its velocity vector parameterized as $$ \vec{v} = \vec{e} \,v $$ where $\vec{e}$ is the tangent vector and $v$ is the speed at that instant. This is kind of obvious. But you use the above to find the tangent vector if you know that radial vector $\vec{r}$. Use $\vec{v} = \frac{{\rm d}}{{\rm d}t} \vec{r} = \vec{e} ...


1

But why must $d\hat{u}$ be orthogonal to $\vec{\Omega}$ too (i.e. be tangential to a circle orthogonal to $\vec{\Omega})$? To get such a precession there must be a clockwise torque in the plane of the screen acting on the system which means that the torque vector must be pointing into the screen. That torque produces a change in the angular momentum ...


1

I think sections 4.1.2 and 4.1.3 of this lecture on dynamics explains it by looking at each component separately. Since $\frac{{\rm d}}{{\rm d}t} \sin \theta = \dot{\theta} \cos\theta$ and $\frac{{\rm d}}{{\rm d}t} \cos \theta = -\dot{\theta} \sin\theta$ the components of ${\rm d}u$ are perpendicular to $u$. Our first step is to choose cartesian axes, ...


1

All you have shown is that the instantaneous velocity of the particle at point $D$ is can be equated to the translation of an arbitrary axis together with a rotation about that axis. The statement "while $\vec \omega$ is independent from the chosen rotation axis, $v$ depends on it" refers to all the particles in the body. So the all have the same ...


0

The velocity of any point is always same.



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