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4

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 ...

0

$\alpha = d\omega /dt = (3120-1200)/16=120 rpm/s$ Number of revolutions, $$\theta = \omega_0t + \frac12\alpha t^2$$ Substitute, $\omega_0 = 40\pi$ and $\alpha$ from above.

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 ...

0

The moment of inertia of a hollow sphere would be higher than a solid sphere of equal radius, only if the unmentioned assumption (same mass) is true! This is typically untrue, because of another assumption, that the hollow and solid spheres (of equal radius) are made of the same density material. If they are made of the same density material, there is no ...

0

If you consider the sphere as two hemispheres, then the centre of gravity of each is 3/8r from the centre. (http://my.safaribooksonline.com/book/mechanical-engineering/9789332503489/6-centroid-and-moment-of-inertia/chap6_sub14_xhtml) The hollow sphere, when considered as two thin shelled hollow hemispheres has a c. of g. at r/2 from the centre. ...

0

Because perpendicular force x radius is a measure of angular momentum, and all the AM of different point masses added up to give the total effective AM. So the larger the radius of the masses the better, in this case we can arrange all of them to reside as point masses at the edge of the circle (because rotation is the motion of action).

0

The question is not why - that is misleading. It is when as under certain conditions only is this true. The moment of inertia is proportional to the mass and distribution of that mass about the axis.

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 ...

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 ...

32

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:

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 ...

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 ...

0

Ok.. I hope I understood it. Force of gravity is uniformly acting on the rolling body at every point of the body while it is rolling down. Suppose we push a rigid box on a surface then our force is acting on every point of it. The friction force will be opposite of its acceleration. Now for a rolling body it will depend on how we are moving the body. We can ...

0

First of all friction force is applied by the surface of the plane where the object is moving. when we walk on a surface we push the surface backwards as a result due to friction the surface push us which is the frictional force. It is in the direction we are walking. Now an object which is rolling with acceleration is rolling with increasing velocity. It ...

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 ...

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} ...

0

Here's how I picture this, w/out any calculations. Imagine the spaceship consisting of two massive objects, "front" and "rear" ones, connected with a long massless rod. They follow the same trajectory before and after interaction with the planet, with the "rear" one always a little behind the "front" one. The forces transmitted through the rod are always ...

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