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Conservation of linear vs. angular momentum in two similar cases

Force at the pivot don't affected the linear momentum. Lets look at the equations: for the bullet mass, $~m$ $$ m\,(v_m-v_0)=\lambda\tag 1$$ for the bob mass, $~M~$ $$ M\,v_M=-\lambda\tag 2$$ and for ...
Eli's user avatar
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1 vote

Why is the moment of inertia the rotational analog for mass and not inertia?

There are two aspects of mass. One is inertia, defined by Newton's second law, F=ma. The other is gravitational mass, defined by Newton's law of gravity, F=Gm1m2/r2. They are the same in relativity, ...
Alien from future's user avatar
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Why is the moment of inertia the rotational analog for mass and not inertia?

The equation for linear momentum is $m \times v$. THe equation for angular momentum is $i \times \omega$. Notice the similarity between the two form. The measure of the inertia in the linear case is ...
KDP's user avatar
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Would a homogeneous rod on a fulcrum oscillate or remain at rest after displacement?

As can be seen in the above diagram, part (triangle defined by HCK) of the green half of the balance bar that was on the left when the bar was in equilibrium, is now on the right making the right half ...
KDP's user avatar
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1 vote

Why is the moment of inertia the rotational analog for mass and not inertia?

but in my textbook the definitions of inertia and moment of inertia are very close to one another They are. The term "moment" is synonymous with the term "torque" which is is ...
Bob D's user avatar
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0 votes

Would a homogeneous rod on a fulcrum oscillate or remain at rest after displacement?

If the rod were perfectly homogenous, infinitely thin, placed on the fulcrum precisely at the midpoint, and the end released without imparting any motion, the rod would theoretically remain at rest in ...
Bob D's user avatar
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0 votes

Would a homogeneous rod on a fulcrum oscillate or remain at rest after displacement?

If the rod has insignificant thickness, is uniform, and is perfectly balanced about the fulcrum when level, then the weights of the parts on either side of the fulcrum will be equal no matter what the ...
gandalf61's user avatar
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0 votes

Conservation of linear vs. angular momentum in two similar cases

*Case 2: The bullet hits at some point along the rod. Here conservation of linear momentum does not apply (it doesn't work) .... It does apply if you expand the system to include the mass of the ...
KDP's user avatar
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1 vote

Conservation of linear vs. angular momentum in two similar cases

It is a valid question, that in case 1 it seems that momentum shouldn't be conserved due to the horizontal force of the pivot point. The simple answer to this question is that, the pivot point does ...
EagerToLearn's user avatar
0 votes
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Mapping between generalized forces and external torque of a rigid body whose rotation is described by quaternion is not unique(?)

$\def \b {\mathbf}$ Starting with the Euler equations $$ \b I\,\b{\dot{\omega}}+\omega\times\,\b I\,\omega=\b\tau_{\rm{E}}\tag 1$$ with $$\omega=2\,\b Q\,\b{\dot{z}}\tag 2$$ where $$\b Q= \left[ \...
Eli's user avatar
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0 votes
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Tendon excursion method application/alternative for joint represented by quaternion in biomechanics

The answer is actually in the same paper I cited. The mapping between generalized forces and spatial forces is derived using virtual work: $F_{Q} = 2G^T T'$ where $F_Q$ is a $4 \times 1$ vector of ...
blenderman's user avatar
3 votes

Why can we treat a ball as a point mass to calculate torque?

The torques can be calculated in respect to any point, and it generally requires taking integrals. However, there are some basic tricks to simplify life. One of them is choosing the rotation axis at ...
Roger V.'s user avatar
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11 votes
Accepted

Why can we treat a ball as a point mass to calculate torque?

Calculating the torque on a rigid body w.r.t to the point $\vec 0$ (WLOG) with gravity pointing in a constant direction $\hat n$ is accomplished by integrating over the rigid body, with each ...
jwimberley's user avatar
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0 votes

What happens if we prevent gyroscope to precess?

Start with no spinning mass and think of a tilting plane as a seesaw, not mounted to the ground, but when you tilt it, it has a tilt axis running through the center that is the fulcrum for the seesaw. ...
jnhrtmn's user avatar
0 votes

Resolution of Ehrenfest paradox using only special relativity

Let's say we have a turntable where any point on the perimeter has a instantaneous tangential velocity such that the gamma factor is 2. A ruler placed on the perimeter will length contract to half its ...
KDP's user avatar
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2 votes
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Resolution of Ehrenfest paradox using only special relativity

Instead of a solid disk, we might as well think of a circular train traveling on a circular track. Alice sits in the station. Bob is riding on the train. The Question: The track is shorter in Bob's (...
WillO's user avatar
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-1 votes

Does a rocket moving in a circle expel exhaust at a greater velocity?

Accordingly, must not the exhaust now obtain the entirety of the spent energy? Yes. From the rocket's frame of reference, would the exhaust gases be perceived as exiting at a greater velocity than ...
BowlOfRed's user avatar
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1 vote

Does a rocket moving in a circle expel exhaust at a greater velocity?

Energy is frame dependent. In a inertial frame that is momentarily comoving with the rocket, its kinetic energy is zero, and the kinetic energy of the gas is the same for both cases, if we suppose the ...
Claudio Saspinski's user avatar
8 votes

Does a rocket moving in a circle expel exhaust at a greater velocity?

This is an interesting question, because when it is moving in a circle, the magnitude of its tangential velocity is constant and its angular velocity is also constant. Therefore the total kinetic ...
KDP's user avatar
  • 6,102
5 votes

Does a rocket moving in a circle expel exhaust at a greater velocity?

The energy of the fuel does become the kinetic energy of the exhaust. Let us ignore the fact that the rocket becomes lighter as it burns fuel, and that it will eventually run out of fuel. The rocket ...
mmesser314's user avatar
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3 votes

Is angular velocity about any point in a rigid body always the same?

Doubt 1 It is measured about the axis of rotation. Doubt 2 Yes. For a rigid body all points will complete one revolution, $2\pi$ radians, about any axis of rotation in exactly the same amount of ...
Bob D's user avatar
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2 votes

Is angular velocity about any point in a rigid body always the same?

Each part of the rotating disk will have the same angular velocity. However, they will not all have the same tangential velocity (which will increase as you move further away from the axis of rotation)...
ABetheGammow's user avatar
2 votes

Is it possible to reproduce tennis racket theorem instability with a gymbal under earth gravitational field?

I'm posting a new answer for the following reason: a totally different way of implementing a Dzhanibekov effect demonstration occurred to me. Manufacture an object that is spherical on the outside, ...
Cleonis's user avatar
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0 votes

Can the motion of a rigid body be decomposed into translational and rotational motion about ANY point?

A rigid body $B$, by definition, is such that, for every point $O\in B$ there is a triple of orthonormal axes ${\bf e}_1, {\bf e}_2, {\bf e}_3$ centered at $O$ such that the material points $Q$ of ...
Valter Moretti's user avatar
1 vote

Why does moment of inertia stop at 1/2 as solidness of a cylinder increases?

The moment of inertia comes from the definition of angular momentum for a particle $\vec L = \vec r \times \vec p$. In the case of a disk, $\vec r$ is always perpendicular to $\vec p$ for any point, ...
Claudio Saspinski's user avatar
4 votes

Why does moment of inertia stop at 1/2 as solidness of a cylinder increases?

The reason that the moment of inertia gets smaller as the annulus gets less like a hoop and more like a disc is that more of the mass is closer to the axis of rotation. You can make the moment of ...
rob's user avatar
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0 votes

Why does moment of inertia stop at 1/2 as solidness of a cylinder increases?

Suppose you have all the mass $M$ concentrated in a ring of radius $R_2$. Then $I = MR_2^2$. Now spread the mass out between $R_1$ and $R_2$. $I$ is less because some of the mass is closer to the ...
mmesser314's user avatar
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2 votes
Accepted

Spinning a rope conundrum

You are exciting standing "circularly polarized" waves in the rope. If you continue to spin at a faster frequency, you can create even more "nodes" in the rope (stable points of ...
mike1994's user avatar
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2 votes

Is it possible to reproduce tennis racket theorem instability with a gymbal under earth gravitational field?

A gimbal suspension for the purpose of demonstrating the Dzhanibekov effect (when weightlessness is not available) presents special challenges. In a comment I already linked to the video about the ...
Cleonis's user avatar
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1 vote

Why doesn't centrifugal force change instantly?

Looking from above, when any spinning wheel is nudged anticlockwise it always responds by trying to rotate the spin axis so that the axis is vertical and the wheel is rotating anticlockwise around the ...
KDP's user avatar
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6 votes
Accepted

Why doesn't centrifugal force change instantly?

This behavior is not caused by centrifugal force, but rather by conservation of angular momentum. If you take any spinning object, and draw an arrow aligned with the axis of rotation, that arrow ...
RC_23's user avatar
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2 votes

Implicit definition of internal forces through choice of dynamical description of a rigid two mass pendulum

You are using two different theorems as if they were equivalent, but they are not. Both theorems come from studying a system of particles. The theorem you use in rotational dynamics is that $\tau=\...
Pato Galmarini's user avatar
0 votes

Rolling ball on a surface with friction

Adding friction to the model is also easy - the frictional force depends on the normal force through the coefficient of friction, $F_f = \mu N$ You need to be careful to differentiate between static ...
Bob D's user avatar
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0 votes

Rolling ball on a surface with friction

On an inclined plane, problem would be very similar to the ones without any rotation, the only differences would be static friction and torque. In the case of rolling without slipping (case where the ...
Alexander Djurovich's user avatar
1 vote

Do off-centre forces create additional energy?

There is a standard demonstration in which an off-centre force is provided by a string wrapped around a cylinder on a air-table. Another string is connected to the c-of-mass of an otherwise ...
mike stone's user avatar
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1 vote
Accepted

Coriolis force on a particle moving on sphere

Considering only Earth rotation around its axis with angular velocity $\boldsymbol{\Omega}_E = \Omega_E \mathbf{\hat{z}}$, you're right that at the equator there is no Coriolis force on a point that ...
basics's user avatar
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2 votes
Accepted

Do off-centre forces create additional energy?

If the line of action of the force is not through the centre of mass then body can be thought of as being acted on by a force whose line of action is through the centre of mass which produces a linear ...
Farcher's user avatar
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0 votes

Motion of bowling ball

Here are a few items that you might find useful: In terms of a reference, you might find reviewing something like Schaum's Theoretical Mechanics helpful: https://archive.org/details/...
ad2004's user avatar
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