# Tag Info

1

If you know velocity, you also know acceleration. If you know acceleration, you can set up Newton's second law and all is known except the push on the track. This push is missing in your statement about the centripetal force, as @lucas comments.

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Point C is sure instant centre of rotation for both gears, otherwise they would get teeth broken if any relative slide to each other. Analogy is a wheel on a road having instant centre of rotation at the bottom point thus velocity of top point is twice more than of the car. As far as I understand, confusion point is that first you calculate VO in respect ...

1

Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation. \begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = ... 1 As you noticed, if we use Euler-Lagrange equation on L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy we get\ddot x=0\ddot y = -g Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations? The point is that system only has ...

3

It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ...

0

It is vector because it has a magnitude and a direction. Also typically the components of $\vec{\omega}$ are evaluated based on an inetial coordinate frame and thus only represent the motion of the body and not of the measuring frame. Rotational speed $\omega = \| \vec{\omega} \|$ Direction of rotation $\hat{z} = \frac{\vec{\omega}}{\omega}$ ...

0

Of course it can. Angular momentum (the rotational analog to linear momentum) can be expressed as an axial vector (sometimes called pseudovector), which means a quantity that is similar to a regular vector, except for the fact that it changes sign under improper rotations (such as reflections). I suppose you're refering by "wobble" to nutation, i. e., a ...

0

It is $T^{-1}$. Consider a rod of length $l$, marked at $l/4$, $l/2$ and $3l/4$, and let it rotate with angular velocity $\omega$ about the centre ($l/2$) point. Now quite clearly the end points are moving twice as fast -- they cover twice the distance per unit time -- as the points marked $l/4$ and $3l/4$, so the dimensions can not be $L/T$, as the whole ...

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Your are mistaken. One way of giving the angular velocity is $\omega=\frac{|v| \sin{\theta}}{|r|}$ which gives $\frac{\frac{L}{T}}{L}=\frac{1}{T}$. We are talking about a change in angle over time, the spatial dimension is given by the $r$ radial distance from the origin without which the angular velocity has no meaning. I always recommend writing all ...

1

Indeed the Lagrangian is independent of $\phi$. However, the partial derivative w.r.t $\dot{\phi}$, $$\frac{\partial L}{\partial\dot{\phi}}=mr^2\dot{\phi},$$ contains $r$ and $\dot{\phi}$, both depending on time $t$. Therefore, you need product rule to compute the total time derivative. That's why you have two terms. They come ...

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 ...

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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 $X = \frac {mg}{3}$. ...

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