# Tag Info

53

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 ...

33

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 ...

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 ...

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} ...

6

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.

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

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

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 ...

5

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 ...

4

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. ...

4

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 ...

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 ...

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

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.

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

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

Let the cone lie on the $\hat{X}\wedge \hat{Y}$ plane (z=0) and let the $z$ axis pierce this plane at the cone's apex. If the cone's half angle is $\alpha$, then its axis of symmetry as a function of time is defined by the vector $$A(t)=\cos\alpha \left(\cos(\omega_0\,t) \hat{X} + \sin(\omega_0\,t) \hat{Y}\right)+\sin\alpha \hat{Z}$$ where $\omega_0 = ... 3 I'm outlining this and stating the final result so that the OP gets the fun of figuring this out themselves. Future responders, please don't work this out All you have to do is allow$\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 The wikipedia link does an adequate job of explaining this. You can find a very mathematical discussion in this partial duplicate question, but here is my simple take on it. The angular displacement in three dimensions does have a vector nature in the sense of having both a magnitude (the angle through which you turn something) and a direction (the axis ... 2 Angular displacement is define by change in angle i.e.${\Delta \theta}=({\theta}_1 -{\theta}_2)$, where$\theta$is taken as positive in anti-clock wise. Hence angular displacement${\Delta \theta}$has both magnitude and direction. But let when a particle rotate from a point A in$\theta$angle in anti-clock wise to point B and then back through$\theta$, ... 2 You've almost got it. You would want to express everything in terms of just one variable, otherwise you run into trouble. Right now you have both$\omega$and$\theta$, but there is a relationship between them... However, since you are asked for an expression in$\theta$, it is simpler to use conservation of energy - you don't actually need to know the ... 2 Yes, this is certainly true. Mass is defined by$m^2=E^2-p^2$(in units with$c=1$), where$(E,p)$is the momentum four-vector built out of the mass-energy and momentum. (This defines what's known as invariant mass, as opposed to "relativistic mass.") Mass as defined in this way is not additive, and depends on the motion of the particles within a system. As ... 2 Welcome to the exchange. When you ask "would it be easier to rotate", there are two different answers. If you mean, would it be easier to accelerate the wheel to some predetermined speed, the answer is yes. If you mean would it be easier to maintain that speed, the answer is no. For a rotating body, you need to learn about angular momentum, and angular ... 2 The earth rotates on its own axis, revolves around the center of mass between the earth and moon, and revolves around the sun. If the earthed stopped rotating around its axis, there would obviously be no Coriolis effect and the climate would be significantly different. No more Hurricanes or rotating air masses. Only temperature differences would move air ... 2 Static friction force arises whenever there is interaction between two bodies by direct contact (touch). There need not be any mutual motion between the bodies. This friction force is necessary to explain why the bodies around us maintain their position so reliably. Without friction forces, there would be nothing opposing their mutual motion and the world ... 2 This is not a dumb question: millions if not billions of people have been taught, incorrectly, that a bicycle's stability originates in the gyroscopic angular momentum of the spinning wheels. This is false, and among other things, would make it near-impossible to turn a bike if it were true. In fact, a bike's stability derives from the momentum vector of ... 2 A day is currently about 86400.002 seconds long. If we could just increase the Earth's rotation rate by a mere 2 milliseconds per day we would get rid of the need for those pesky leap seconds. No problem! We only need something that rotates with an angular momentum of 1.4×1026 joule-seconds about an axis pointing due south. One way to do this would be to ... 2 The Earth is currently rotating at one revolution per sidereal day. Converting to radians, this is an angular velocity of$\omega_0 = \frac {2\pi}{\text{sidereal day}}$. You want to have 360 solar days per year, or 361 sidereal days per year. That means a rotation rate of one revolution per 1/361 tropical year, or an angular velocity of$\omega_1 = \frac ...

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