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50

A hollow sphere will have a much larger moment of inertia than a uniform sphere of the same size and the same mass. If this seems counterintuitive, you probably carry a mental image of creating the hollow sphere by removing internal mass from the uniform sphere. This is an incorrect image, as such a process would create a hollow sphere of much lighter ...


25

The key is... the closest the mass to the axis of rotation, the easiest to add angular velocity to the body. For instance a figure skater rotates faster when she puts her limbs closer to her body. Let's see how it works from a more intuitive fashion: For instance, in the figure bellow, trying to lift up the table (A) would be easier compared with the ...


17

The moment of inertia of a body about an axis is a measure of how far the mass is distributed from that point. For a solid sphere of mass $m$, radius $r$, you have the mass distributed continuously from the center to the radius. However, for a hollow sphere of mass $m$, inner radius $r_i$ and outer radius the same as before, $r$, you have all the mass ...


11

The system needs to conserve momentum. In both cases, the momentum is whatever m*v is for the bullet. Since it's the same in both cases, the bullet and block have the same vertical velocity. Mechanical energy is not conserved. The reason the block hit on the side has more kinetic energy is that the bullet converted less of its kinetic energy into heat upon ...


9

The equation you are referring to is the expression for the moment of inertia of a point particle of mass $m$ at a distance $R$ away from some axis. This is expression is really the definition of the moment of inertia for a point mass, so the question becomes "where does this definition come from, and why is it useful?" Well for simplicity, suppose that ...


7

There is a identity for the derivative of the cross-product of two vector functions $\mathbf A(t)$ and $\mathbf B(t)$; \begin{align} \frac{d}{dt} (\mathbf A \times \mathbf B) = \frac{d\mathbf A}{dt}\times \mathbf B + \mathbf A\times \frac{d\mathbf B}{dt} \end{align} Using this rule with the computation you're considering, we obtain \begin{align} ...


5

No, these building are still tiny compared to earth's crust mass distribution. One would need to build whole mountain ranges to detect changes in earth gravity field with high precision instruments. And even those wouldn't changed earth orbit measurably because even a mountain range is tiny compared to the mass of the whole earth. However mountain ranges ...


5

When you compute the energy of a rotating object you either consider the tangential velocity or the angular velocity. The two expressions are two different ways to look at the same observable, they both lead exactly to the same result and summing them is a mistake. For instance for a point like mass on a straight line we have: $$E_l = \frac{1}{2}m v^2$$ ...


4

If you apply the same force for the same period of time, the linear velocity of the body will be the same in both cases, assuming the body is unconstrained. However, having applied the same force for the same amount of time does not mean that the same amount of energy has been transferred. The energy, or the work done by the force, is the force times the ...


4

Assume that you jump straight up, standing on the equator. As soon as your feet leave the ground, you are in a highly elliptical orbit around the center of the earth. At that point you have the same angular velocity as the point you jump from. As you rise toward your one and only apogee, conservation of angular momentum requires that your angular velocity ...


4

The vector product of a vector $\vec{a}$ with itself is alwals zero: $\vec{a} \times \vec{a} = 0$ For two smooth vector-valued functions $\vec{a},\vec{b} \colon \mathbb{R} \to \mathbb{R}^3$ the product rule holds: $$ \frac{d}{dt} (\vec{a} \times \vec{b}) = \frac{d}{dt} \vec{a} \times \vec{b} + \vec{a} \times \frac{d}{dt} \vec{b} $$ You can see this for ...


4

No, it does not gain energy. The confusion arises because there's a force that does no work. If the car moves a distance $d \vec l = \vec v dt$, then the work done during that time $dt$ is $dW_{rope} = \vec F_{rope}\cdot d \vec l= 0$. This follows because $ \vec F_{rope}$ and $d \vec l$ are always perpendicular to each other (draw and check this!) which ...


3

The answer to that is because the moment of inertia is not the same for the solid cylinder than for the hollow one. As you write the formula for the moment of inertia, it depends on the distribution of the mass. The further away the mass is from the rotation axis, the more contributes to the moment of inertia (as in distance squared $r^2$). So, since the ...


3

For a ball rolling down a ramp, using the KE and PE, you can find that it will be $$ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $$ Use $v = r\omega$ and $I = 2/5 \space mr^2 $ and you get $a = 5/7 \space g \space sin\theta$ ( $\theta$ is the angle of incline). Do not forget to include the errors in your measurements both in distance and in time, so ...


3

The total velocity will be the sum of the translational and rotational velocities. Thus $$ \mathbf{v}_\text{net}=\mathbf{v}_\text{COM}+\mathbf{\omega}\times \mathbf r, $$ where $\mathbf r$ is the vector from the center of mass to the vertex, and $\mathbf v_\text{COM}$ is the center of mass velocity.


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

A rigid body has 6 configuration degrees of freedom because its most general configuration can be obtained by translating (3 degrees of freedom) and rotating (3 degrees of freedom) its initial configuration. A mathy way of saying this is that its configuration manifold is $\mathbb R^3\times \mathrm{SO}(3)$. However, you are right that the phase space of a ...


3

$\newcommand{\xhat}[0]{\hat{x}}$ $\newcommand{\yhat}[0]{\hat{y}}$ $\newcommand{\zhat}[0]{\hat{z}}$ $\renewcommand{\c}{\cos \theta}$ $\newcommand{\s}{\sin \theta}$ $\newcommand{\fce}{F_\textrm{cent}}$ $\newcommand{\fco}{F_\textrm{cor}}$ $\newcommand{\vfce}{\vec{F}_\textrm{cent}}$ $\newcommand{\vfco}{\vec{F}_\textrm{cor}}$I saw a question I wanted to answer ...


2

Your angular velocity vector is $$ \vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|} $$ where $\vec{r}_A = (0,0.2,0.12)$, $\vec{r}_D = (0.3,0,0)$, $\vec{r}_B = (0.3,0.2,0.12) $ in meters and $\Omega = 90\;{\rm rad/s}$. Your velocity kinematics is $$ \vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A ) $$ And acceleration ...


2

It looks arbitrary because the concept of Inertia is defined that way. If someone can digest $F=ma$ (linear), the same should be followed for $\tau=I\alpha$ (angular) Moment of Inertia is just the rotational substituent for the inertia coefficient (mass) in linear motion. It's not some kind of arbitrary value. It's actually given by $I=Kmr^2$. The $mr^2$ ...


2

A simple example would be two baseballs of mass M connected by a 1-meter stiff bar, placed on tees 1 meter apart. You hit one baseball with a bat, but not the other one. This imparts a velocity V to the one you hit, and velocity 0 to the one you didn't hit. So immediately after hitting, the total momentum is $MV$, and the kinetic energy is $MV^2/2$. If ...


2

Here, in the above picture $M \ge m$. Note that this is the most general case. We can have $M = m$ and the angle $\theta$ can vary anywhere between $[0;\cfrac{\pi}{2}]$ (Actually, the most general case would have been to take 4 different masses but we will be going out of the bandwidth of your problem, and it would be a pointless discussion.) Now, ...


2

The expression you quote is for a ideal monatomic gas, and we get $C_v = 3/2$ for the three degrees of freedom. For ideal diatomic gases we do indeed have to count rotational degrees of freedom and we get $C_v = 5/2$. See the Wikipedia article on ideal gases for more info.


2

So when we see an object rotating, its state of rotation is totally relative, as it happens for many other physical quantities... Is that correct? The state of rotation is not $totally$ relative; for example, the angular velocity of rotation is the same for all points of reference. It is true that you can use different point of reference for the ...


2

usually linear velocity is the velocity of a point rotating around the axis of rotation given by $$ \vec v = \vec{\omega} \times\vec{r} $$ when object has no translational motion but if the object has both translational and rotational motion then $\vec v$ will be measured from Center-of-momentum frame. in the frame where the C.M is translating at velocity ...


2

I think there are 2 main sources of confusion: First, because of gravity, extending your arms feels like work. We're only interested in the radial movement, though, and in this direction, the skater's arms are pulled by the centrifugal force (in the long tradition of spherical cows in vacuum, we could replace the figure skater with two beads on a spinning ...


2

I do not know anything about Cayley-Klein parameters, sorry, but the rest of you question is easy. We start noticing that it is always possible to suppose that ${\bf C}$ is directed along $z$, just fixing our reference frame suitably. So we can write ${\bf C}= c{\bf e}_z$ whit $c>0$ (otherwise everything is trivial and the solution is ${\bf m}(t)= {\bf ...


2

If you assume there is no friction, the car can not turn. Why? Refer to Newton's first law: The velocity of an object does not change, unless acted upon by an external force, when viewed from an inertial frame of reference. Notice we say velocity, not speed. Since velocity includes direction, turning the car means changing the velocity. Since the only ...


1

Suppose we have a particle moving with a velocity $\vec{v}_1$ and a object which has no linear velocity, but is rotating with angular velocity $\vec{\omega}$. Now suppose we look at a point at a displacement $\vec{r}$ from the axis of rotation of this rotating velocity. The velocity of this point is $\vec{v}_2 = \vec{\omega} \times \vec{r}$. Now the ...


1

The angular momentum of a massive sun may cause the freely falling spaceship to start spinning in the direction of the sun's angular momentum for an effect of frame dragging. You can take a look at the Kerr metric which describes the behaviour of the spacetime near a massive spinning object. If you're not familiar with general relativity it could be ...



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