# Tag Info

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, ...

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

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 ...

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

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

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 ...

3

As Manishearth says, for engines with more than one cylinder the firing of the other cylinders rotates the crankshaft. However, as any fan of vintage motorcycles will know, you can have four stroke engines with a single cylinder. In this case the engine has a heavy flywheel attached to the crankshaft and the momentum of the flywheel keeps the crankshaft ...

3

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 ...

3

COMPLETE REWRITE: For small jumps, the answer is 122micrometers*s^3, where s is the time of the jump in seconds. I used numerical methods in Mathematica. Can someone improve on and/or verify this solution? Consider this diagram: I then ran the following: (* Earth's radius in meters *) r = 40000000/2/Pi (* seconds in day *) d = 86400 (* ...

3


3

Let $\vec r_0(t)$ denote the point around which the object is rotating and $\vec r(t)$ the position of the object. Then the fact that the particle is rotating around the point $\vec r_0(t)$ can be formalized by the mathematical statement that $$\vec r(t) - \vec r_0(t) = R(t) \vec c$$ for some constant vector $\vec c$ and time-dependent rotation $R(t)$. ...

3

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 ...

3

The total velocity will be the sum of the translational and rotational velocities. Thus $$\mathbf{v}_\text{net}=\mathbf{v}_\text{COM}+\mathbf{\omega}\times \mathbf r,$$ where $\mathbf r$ is the vector from the center of mass to the vertex, and $\mathbf v_\text{COM}$ is the center of mass velocity.

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 ...

2

You are right; the magnitudes of the translational and rotational kinetic energies are the same. In general, the kinetic energy of a mass distribution can be written $$K = K_{cm} + K_{rel}$$ $K$ is the total kinetic energy. $K_{cm}$ is the kinetic energy of a point particle whose mass is equal to the total mass of the distribution, if the point ...

2

A two-wheeled robot cannot move in the direction parallel to the axis of its wheels. Therefore, it cannot move in an arbitrary direction independent of its rotation; it is not holonomic. This implies that your robot cannot move in a straight line while spinning about its center (as a weightless/frictionless body would). However, it sounds like you are OK ...

2

This is actually fairly straightforward, at least based on what you've described. You have a left wheel which moves at an angular velocity $\omega_L$ and a right wheel which moves at an angular velocity $\omega_R$. The distance traveled by the edge of each wheel in a time $t$ is $r\omega t$, where $r$ is the radius of the wheel, and accordingly $r\omega t$ ...

2

The problem is exactly analogous to a very famous problem which is presnt in any good Rotational book, Whats happening here is ball has a velocity of its centre of mass,Vcm, and an angular velocity. Now when we put the ball on a friction ON surface, the force of friction provides a torque and an acceleration to the centre of mass, this happens till the ...

2

The definition of rotational kinetic energy is $$E_\text{rot}{}_{(i)} = \frac{1}{2} J_i \omega_i^2 = \frac{\;\; L_i^2}{2J_i}$$ where $J_i$ is moment of inertia, $\omega_i$ is angular velocity and $L_i = J_i\omega_i$ is angular momentum of the particle. If you select $\vec{r}_\text{cm}$ as the center of rotation these values can be calculated for each ...

Only top voted, non community-wiki answers of a minimum length are eligible