# 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

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

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

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 ...

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 ...

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

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


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

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 ...

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 ...

2

Your angular velocity vector is $$\vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|}$$ where $\vec{r}_A = (0,0.2,0.12)$, $\vec{r}_D = (0.3,0,0)$, $\vec{r}_B = (0.3,0.2,0.12)$ in meters and $\Omega = 90\;{\rm rad/s}$. Your velocity kinematics is $$\vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A )$$ And acceleration ...

2

It does, but changes are so miniscule that they don't really count. Even if everyone in the world started running in the same direction (East, for example), their total angular momentum would be on the order of $10^{18} \ \text{N m s}$, in comparison to the earth, which is on the order of $10^{33} \ \text{N m s}$. We would change the earth's rotation by ...

2

It looks arbitrary because the concept of Inertia is defined that way. If someone can digest $F=ma$ (linear), the same should be followed for $\tau=I\alpha$ (angular) Moment of Inertia is just the rotational substituent for the inertia coefficient (mass) in linear motion. It's not some kind of arbitrary value. It's actually given by $I=Kmr^2$. The $mr^2$ ...

2

A simple example would be two baseballs of mass M connected by a 1-meter stiff bar, placed on tees 1 meter apart. You hit one baseball with a bat, but not the other one. This imparts a velocity V to the one you hit, and velocity 0 to the one you didn't hit. So immediately after hitting, the total momentum is $MV$, and the kinetic energy is $MV^2/2$. If ...

2

The expression you quote is for a ideal monatomic gas, and we get $C_v = 3/2$ for the three degrees of freedom. For ideal diatomic gases we do indeed have to count rotational degrees of freedom and we get $C_v = 5/2$. See the Wikipedia article on ideal gases for more info.

2

Here, in the above picture $M \ge m$. Note that this is the most general case. We can have $M = m$ and the angle $\theta$ can vary anywhere between $[0;\cfrac{\pi}{2}]$ (Actually, the most general case would have been to take 4 different masses but we will be going out of the bandwidth of your problem, and it would be a pointless discussion.) Now, ...

2

usually linear velocity is the velocity of a point rotating around the axis of rotation given by $$\vec v = \vec{\omega} \times\vec{r}$$ when object has no translational motion but if the object has both translational and rotational motion then $\vec v$ will be measured from Center-of-momentum frame. in the frame where the C.M is translating at velocity ...

1

The energy formula $$\tag{39.11} E~=~\frac{1}{2}mv^2 -\frac{1}{2} m({\bf \Omega} \times {\bf r})^2 + U$$ in Ref. 1 (of a point particle, as seen in a rotating reference frame $K$) consists of three terms: Kinetic energy: $\frac{1}{2}mv^2$. Centrifugal potential energy: $-\frac{1}{2} m({\bf \Omega} \times {\bf r})^2$. Other potential energies $U$. ...

1

I see the problem with your equation now. When differentiating $\vec r=r\cos\theta\vec i+y\vec j$, you have considered $r$ to be constant, which is wrong. $r$ is given by $$r=\sqrt{l^2+y^2}$$ where $l$ is the side-length of the square. So $r$ will change with $y$, and you'll have to differentiate $r$ too. This is where the math gets pretty ugly and ...

1

$\Delta \theta$ doesn't necessarily need to be a scalar. Check this Wikipedia article about angular displacement in 3D. It can be denoted as a vector, having magnitude equal to the radians covered, and direction according to the Right Hand Thumb rule. Saying that, there is no difference between $\Delta\theta$ and $\delta\vec\theta$. Maybe you regard ...

1

The disc is rotating, so every point on the disk is moving, thus have the kinetic energy, consider a small point $P$ with mass of $m$ on the disk with a distance from the center of rotation of $r$, apparently, the kinetic energy of this point is $E = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2r^2$ Now, the disk consists lots of these "points", so the total ...

1

That line will get Coriolis acceleration $$\vec{a} = -2 \vec{\Omega} \times \vec{v}$$ ($\Omega$ is the angular speed of the earth's rotation, with a direction pointing into the ground from the view of the south pole). As it's going across the pole, there's a right angle between $\Omega$ and $v$ and the absolute value will be simply $$a = 2\Omega v$$ and the ...

1

When the player hits the ball with top spin, it makes the ball, well, spin. By spinning, the ball will modify the airflow around itself and thus create an air pressure profile which will deflect the ball : this is the Magnus effect. So by applying top spin on the ball the way tennis players do, the ball is rotating in the direction of the trajectory. This ...

1

Because you've not employed the right angles yet. The velocity of something moving in a circle is necessarily tangent to the circle at whatever point it lies. That property means that the velocity vector is at a right angle to the radius vector. The radii vectors are AB and AC, and their corresponding velocity vectors are v1 and v2. We know that the ...

1

Let a system consist of particles with positions $\mathbf x_i$ as measured in some inertial frame, and let $\mathbf x'$ denote the position of the center of mass of the system, then if we define the center of mass positions by $$\mathbf x'_i = \mathbf x_i -\mathbf x'$$ then we have $$\mathbf x = \mathbf x' + \mathbf x_i'$$ And the angular momentum of ...

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