# Tag Info

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 ...

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 ...

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 ...

1

Suppose you have a satellite of mass $m$ at a distance $r$. If we assume the satellite is small enough to behave as a point mass the moment of inertia of the satellite is: $$I = m r^2$$ so its kinetic energy is: $$E = \tfrac{1}{2} I w^2 = \tfrac{1}{2} m r^2 \omega^2 \tag{1}$$ But for a body moving in a circle of radius $r$ at an angular velocity ...

1

Interesting question. Normal forces are normal: perpendicular to the surface. So where does the centripetal acceleration come from? The only place I can see it coming from is friction between the earth and my shoes. A simple calculation shows that the centripetal acceleration at mid latitudes is about 20 mm/sec${}^2$. Consider a hanging plumb bob. The ...

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