# Tag Info

## Hot answers tagged rotational-kinematics

51

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 ...

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:

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 ...

12

In principle, yes, but the effects are almost completely negligible. As objects on the surface of the earth move around, the Earth's moment of inertia changes by minute amounts, and this affects its rotation. However, performing an order of magnitude estimate on the ratio of the contribution to the moment of inertia of a person-sized object on the equator ...

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

This is a note on why angular velocities are vectors, to complement Matt and David's excellent explanations of why rotations are not. When we say something has a certain angular velocity $\vec{\omega_1}$, we mean that each part of the thing has a position-dependent velocity $\vec{v_1}(\vec{r}) = \vec{\omega_1} \times \vec{r}$. We might consider another ...

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 are actually several different ways to interpret that question, depending on what you mean by "vector" and "rotation". But here's a sense that I've often wondered about myself: in introductory physics, the velocity vector is defined as the time derivative of the position vector (relative to some fixed point). Why is the same not true of angular ...

7

I'll tackle your questions in reverse: 3. The contact point is stationary because the wheel is not slipping. This happens when the force of static friction is able to counter the force of the wheel on the ground. This is what you want for controllable transport. If the wheel starts slipping (because of low friction) that's a skid and you are no longer able ...

7

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun. However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, ...

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

One way to explain it that makes sense to me is that the backwards spin on the ball is a force pushing the ball toward you. The one time applied force of your finger that creates the backward spin also pushes the ball forward, but the backspin stays almost constant. At the moment you spin the ball, the forward force is greater then that of the backspin, so ...

6

There is no gravitational waves for a uniformly rotating axially symmetric body, because the metric doesn't depend on time. First of all, let me cite Landau, Lifshitz, The classical theory of fields, §88 The constant gravitational field: However, for the field produced by a body to be a constant, it is not necessary for the body to be at rest. Thus the ...

6

I agree with the previous answer. Angular momentum, something the earth has because of its rotation about its axis, can only be changed when an external torque (twisting motion) is applied to the earth. As far as I know, there are two ways in which this can happen. If there was friction between the earth's surface and space, then that would slow down the ...

6

Defining properties of vectors are that you can add them and multiply them by constants. These both make sense for angular velocities. On the other hand, adding rotations doesn't make sense. What you can do with two rotations is compose them: first rotate one way, then rotate another. This operation doesn't look like addition of any sort. For one thing, it ...

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

If there is no external torque, then the angular momentum of the system will be conserved. This seems to be the case in the system you're describing, assuming no frictional forces. The spring will contribute to the moment of inertia of the system as a whole(*), but once you have the whole system rotating with a given angular momentum, it should continue to ...

5

Can you get any further when you know the Taylor expansion of $cos(\Delta \theta)$ $$cos(x) \approx 1 - \frac{1}{2}x^2 + \mathcal{O}(x^4)$$

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

Angular rotation is a vector so at any given instant any rigid body can only be rotating about one axis. If the body is rotating freely in space with no external forces then angular momentum is conserved. If the object is spherically symmetrical like the ball you suggest as an example, then the angular velocity is in the same direction as the angular ...

4

It looks just like rotation around a different axis with a different rotational speed. Specifically, if you set an object to rotate with angular velocity $\vec\omega_1$ and also with angular velocity $\vec\omega_2$, then it's really rotating with angular velocity $\vec\omega_1 + \vec\omega_2$. The direction of the vector $\vec\omega_1 + \vec\omega_2$ is the ...

4

The total linear momentum of a system of particle labeled with $i \in {1\dots n}$ can be found in the microscopic view just by summing the linear momentum of the constituents: $$\vec{P} = \sum \vec{p} = \sum m_i \vec{v}_i$$ Now, writing $M = \sum m_i$ for the total mass, $\vec{X}$ as the position of the center of mass, and $V$ as the velocity of the ...

4

For the person not to slip, there must be a centripetal force of $mv^2/r = m r \omega^2$ towards the centre. Since $v$ varies with $r$ while $\omega$ is fixed ($v=r\omega$), it is probably easier to take the second form, in which case this force has to increase as $r$ increases. This forces comes from friction since there are no other forces in the plane ...

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

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.

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 ...

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