# Tag Info

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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 ...

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Here's an illustration of a uniform sphere and a hollow sphere mid-sections with the same mass, if you better understand these things visually:

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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 ...

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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 ...

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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} ...

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The inertia ellipsoid is computed from an integral about an axis - in other words you rotate the object. This will "smooth out" any symmetries and typically increase the symmetry. Sorry this is a "early morning" intuitive explanation - maybe someone else will give you a more formal answer.

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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: $$E_l = \frac{1}{2}m v^2$$ ...

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The key is the coriolis force. The coriolis force is $F_c = -2m\Omega \times v$. Here $\Omega$ is the rotation of the frame of reference and $v$ is the linear speed of the satellite. If you do the calculations, left as an exercies for the reader, you'll get the missing force. In case 2 the coriolis force is 0, because the velocity $v$ has to be used in ...

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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 ...

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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 ...

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Assuming non friction, we must assume the car turns because it is held by a rope or chain to a point. The turning itself will not take energy from the system, what you have now is a constant speed that is tangent to the circle the car is following at the current car position.

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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 ...

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Angular displacement is an example of what's generally called a pseudovector. This is a quantity that is similar to a regular vector, except for the fact that it behaves differently under improper rotations such as reflections (it gains an additional sign flip). Any quantity which is the cross-product of two polar vectors will generally be a pseudovector. ...

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I have reproduced your calculation. If $\theta$ is the angle from the central vertical axis of the hemisphere to the ball, the tangential downward force is $g \sin \theta = g\frac rR$ The tangential upward force due to rotation is $\omega ^2 r \cos \theta=\omega ^2 r \sqrt {1-\frac {r^2}{R^2}}$ When $\omega \lt \sqrt{\frac gR}$ the downward force is ...

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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 ...

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The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and ...

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I'm outlining this and stating the final result so that the OP gets the fun of figuring this out themselves. Future responders, please don't work this out All you have to do is allow $\omega(t)$ to be a function of time. You'll get extra ${\dot \omega} = \alpha$ terms in your equation, and you'll get a final result that says that $${\vec a} = {\vec ... 3 Potential energy is energy of a position or orientation relative to other positions or orientations. For example, if the hamster wheel is on a table, it will have more gravitational potential energy than if it is on the floor upon which the table is standing. If the hamster wheel is rotationally symmetric about the axis of rotation, and the axis is not ... 3 A rigid body has 6 configuration degrees of freedom because its most general configuration can be obtained by translating (3 degrees of freedom) and rotating (3 degrees of freedom) its initial configuration. A mathy way of saying this is that its configuration manifold is \mathbb R^3\times \mathrm{SO}(3). However, you are right that the phase space of a ... 3 In the basic discussion of angular momentum where something is rotating around a fixed symmetrical axis \vec{L}=\vec{r}\times\vec{p} reduces to \vec{L}=I*\vec{\omega} Like in this animation where each vector is colored appropriately: However angular velocity and angular momentum can have different directions in two cases: If the axis of ... 3 For a ball rolling down a ramp, using the KE and PE, you can find that it will be$$ 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 ...

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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 ...

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Cool question. Let's try to formalize the general concepts presented in the other two responses thus far. Intuitively, the main idea is that the instantaneous axis of rotation is determined by the rotation that takes the object from its configuration at one instant to its configuration at the next instant as opposed to the rotation that takes it from its ...

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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.

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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, ...

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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 ...

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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 ...

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