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It is because the moment of inertia is not a conserved quantity. The statement that an isolated body can't change its position is more precisely the statement that an isolated body cannot change the position of its centre of mass. The position of the centre of mass, ${\bf R}$, is given by: $$ {\bf R} = \frac{1}{M}\sum m_i {\bf r}_i $$ where $M$ is the ...


10

John correctly stated that this is possible because re-configuring our bodies allows us to change our moment of inertia, but not our mass. As the question was about an intuitive explanation, consider adding a series of floating weights to get an analogous situation for translational motion: The astronaut stretches their arms above the head, grabs a weight, ...


6

The rotational energy of a body is given by: $$ E = \tfrac{1}{2}I\omega^2 $$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a uniform sphere the moment of inertia is related to the mass of the sphere, $m$, and the radius of the sphere, $r$, by: $$ I = \frac{2}{5}mr^2 $$ You already have the mass, and you can Google for the ...


5

John Rennie's answer is correct in the special case that angular momentum and angular velocity are parallel to one another. This is not always the case as moment of inertia is a second order tensor. Angular momentum is given by $\boldsymbol L = \boldsymbol{\mathsf{I}} \boldsymbol \omega$. Differentiating this gives the rotational analog of Newton's second ...


4

What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2? The equation you wrote doesn't mention anything about the linear acceleration. Is the center of mass located at its top and not the center? Or is there something else I am missing? The center of mass of the pencil is in the middle, not the top. There is likely something else ...


3

"Equation that is all over the internet"... You started at http://thatsmaths.com/2014/06/26/balancing-a-pencil/ and from there, you linked to http://arxiv.org/pdf/1406.1125v1.pdf which was the source for the former. In the third paragraph of that paper, it states We model the pencil as an inverted simple pendulum with a bob of mass m at one end of ...


3

Even if I ignore wind and the drag forces and only consider the rotation of the earth the bullet will not hit the ground at the same place from where it was projected. There will be Coriolis effect. Coriolis effect: The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame. I suggest this ...


3

It seems helpful to consider an extremely simple scenario. Suppose an astronaut is floating near two balls of lead; in this case the closed system consists of the astronaut together with the balls. She can pull the balls together without changing the momentum or angular momentum of the system. She can then rotate them in the center with almost no change, and ...


3

Moving bodies have inertia which means that they will continue to move at a constant velocity unless acted upon by an external force (this is Newton's first law of motion). Similarly, rotating bodies have a moment of inertia, meaning that they will continue to rotate unless acted upon by an external force (torque). Therefore, torque is only required to ...


2

The equivalent to force is torque, $\tau$. So the second law would be: $$ \tau = I\dot{\omega} = I\ddot{\theta} $$


2

The dynamics of a ball rolling down an incline is interesting. Let's start by figuring out the forces that come into play for the non-slipping case (mass m, radius R, angle of ramp $\theta$): If we consider the motion of the ball as a rotation about point $P$, then the torque is given by $$\Gamma = mgR\sin\theta$$ and the moment of inertia about $P$ is ...


1

Assuming earth mass, procession would be reduced practically to zero because tidal effects would be practically zero. Procession, at least according to this site, has to do with tidal effects and a planet being mailable. A black hole - 2/3rds of an inch in diameter would experience essentially zero tidal effects and it would be far less prone to bulging ...


1

This builds partially off of TZDZ's answer. In order to fully understand the moment of inertia, you need to fully understand the idea of torque. The only sensible way to fully understand torque that I've seen is to first understand it as the derivative of rotational work $W_R$ being done with respect to the angle $\theta$ through which the work is done. So: ...


1

Your intuition is correct: you can't find the centripetal force from an analysis of only forces. The centripetal force is found from an analysis of motion. The problem is that one of the forces varies in order to meet the requirements of the motion: it is a force of constraint. Consider a test particle on the rim of the disk. There are two forces on ...


1

Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that $\mathbf M_O$, $\vec{OP}$ and $\mathbf F$ belong to $\mathbb{R}^3$, as MyUserIsThis did in his comment. If this is not satisfactory to you, you could consider three distinct fields of numbers, say $\mathbb{R}_F$ for forces, ...


1

I will answer the mass moment of inertia tension question. For an infinitesimal clump of mass ${\rm d}m$ located at $\vec{r}$ its effect on the inertia tensor is $${\rm d}{\bf J} = -[\vec{r}\times][\vec{r}\times]{\rm d}m$$ where $[\vec{r}\times]$ is the skew symmetric cross product operator $$ \begin{pmatrix}x\\y\\z \end{pmatrix}\times = ...


1

Integrating your equations (1) and (2) you get that $(\text i) \ \omega_1 = C_1$ a constant in time, while another constant in time. This equality you can re-write as $(\text {ii}) \ \omega_2^2 + \omega_3^2 = C_2^2$, As to the vector $J_{\Omega}\omega$ you know that its components are $\{J_1\omega_1, J_2\omega_2, J_2\omega_3\}$, therefore $(\text ...


1

Of course you can specialise the derivation of the three dimensional Euler equations for your special case. However, unless you are expressly asked to do this, I doubt that your teacher means for you to take this path because it is very much an overkill. The yoyo is spinning in a plane about its axis of symmetry. The inertia tensor is now a simple scalar ...


1

For a rotating ball you should use for energy $$E=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2+mgh$$ where $I$ is the moment of inertia which for a sphere is $$I=\frac{2}{5}mr^2$$ where $r$ is the radius of the sphere, and $\omega$ it's his angular velocity which is related to the velocity of the center of mass via the equation $$\vec{r}\times \vec{\omega}=\vec{v}$$ ...


1

To understand rolling without slipping, first consider the case of rolling only case about the center of mass.In this case, a point on the top rim will have a speed $v=\omega\cdot R$ and a velocity $v=-\omega\cdot r$, at bottom of the rim, as observed by you. However in case of rolling without slipping, We observe the velocity of center of mass is $v$ (only ...



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