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37

The following fact lies at the heart of this and many similar issues with sizes of things: Not all physical quantities scale with the same power of linear size. Some quantities, like mass, go as the cube of your scaling - double every dimension of an animal, and it will weigh eight times as much. Other quantities only go as the square of the scaling. ...


29

The units for torque, as you stated, are Newton-meters. Although this is algebraically the same units as Joules, Joules are generally not appropriate units for torque. Why not? The simple answer is because $$W = \vec F \cdot \vec d$$ where $W$ is the work done, $\vec F$ is the force, $\vec d$ is the displacement, and $\cdot$ indicates the dot product. ...


11

Newton's first law of motion for a point particle states that a particle at rest will stay at rest and a particle in motion will stay in motion unless acted on by an unbalanced force. In other words, if the net force on the particle is zero, then the velocity of the particle will stay constant. Newton's first law of motion for a system of particles states ...


10

Take the right pedal as an example. It uses a right-hand thread, so turning the pedal spindle clockwise (CW) relative to the crank will screw the spindle in, counter-clockwise (CCW) will unscrew it. Say you put the bike in a repair stand, grab the right pedal and gently simulate the motion of someone riding the bike (always keeping the pedal platform ...


10

You can grow arbitrarily large as long as you are essentially flat. For example, one fungus covers several thousand acres; there's a grove of clonal aspen trees that may have higher mass. Scaling in three dimensions is much harder, though. The pressure on the bottom is proportional to the height--eventually that pressure is too great for tissue to ...


9

If you want to prove that $\vec{L}=\vec{r}\times \vec{p}$ is constant with respect to time for a particle in a central force field $\vec F = \phi(r) \vec r$, just show that the angular momentum doesn't change with time, i.e. $\frac{d}{dt}\vec{L}=0$. Using the product rule we get two terms: $\frac{d}{dt}\vec{L}=\frac{d}{dt}(\vec{r}\times \vec{p}) = ...


8

The moment of a vectorfield $\vec{v}$ at a position $\vec{r}$ is equal to $$\vec{r}\times\vec{v}.$$ So torque is simply a special case where the vectorfield we look at is the force field, $\vec{v} = \vec{F}$. Another way of saying this is that torque is the moment of force.


7

When studying angular things - torque, angular velocity, angular momentum, etc. - physicists do a clever thing to avoid having to describe curves. You see, you might be tempted to draw a curved arrow for a torque, indicating that you are twisting something around in a circular-ish way. But then when you try to add two such arrows together, all of a sudden ...


7

The reason we distinguish the two is that torque is vector quantity, where as energy is a scalar quantity. So while we give the magnitude of torque the same units as energy, there is in fact additional information that tells us the direction the torque is applied. UPDATE: As dmckee has pointed out in the comments, to be perfectly corrected torque is a ...


7

There is indeed a term involving the time derivative of the changing coupling between the masses. First, let's derive the equation for a single mass. $$L = \frac{1}{2} I\, \dot{\theta}^2 - V(\theta)$$ $$\frac{\partial L}{\partial \dot{\theta}} = I\, \dot{\theta}$$ $$\frac{\partial L}{\partial \theta} = -\frac{dV}{d\theta} = \tau$$ $$\tau = \frac{d}{dt} ...


7

Yes, electric motors can always deliver the required torque. What really matters is the efficiency of the conversion. Over a long range of speeds, the motors may convert 90 percent of the energy to its mechanical form. You know, it's just some electromagnetic fields that may become arbitrarily stronger - depending on how much energy is sent into them - and ...


7

Since torque is defined as the rate of change of angular momentum, the more fundamental question would be whether angular momentum is a vector in SR. The answer is no, because there is no vector cross product in four dimensions. Angular momentum is a rank-2 tensor.


6

As in the comments, there's certainly something of a convention at work here and it's to do with the "co-incidence" that we live in three spatial dimensions. As in Greg's answer, torque is intimately linked with angular momentum through Euler's second law. That is, torque and angular momentum are about rotational motion. And rotations, in general, are ...


6

Some engineering texts use "moment" and "couple" to talk about forces that tend to rotate an assembly (what physicist mean when they say "torque", but the engineers sometimes have a slightly different meaning for that word). A roughly translation guide is... A "couple" is a pair of opposite forces whose points of action are not co-linear. A couple is ...


6

First of all, choose the reference frame co-moving with the paddle and assume that this reference frame is inertial. This is the key to all ball-and-wall problems. Of course, ignore gravity and air drag. Now we have a spinning ball incident on a stationary surface with friction. Let $\mathbf{V}$ be the vector of the ball's velocity with respect to the ...


6

Ingo, when you consider the couple, you may put one of the "spouses" at the origin, so his torque is $P\times d_0$ for $d_0=0$, so his torque vanishes. Meanwhile, she is located at a nonzero $d$ so her contribution is $P\times d$ and nonzero. Because his torque is zero, it doesn't matter whether you add him or not. The only difference between the whole ...


6

Torque is the informal, practical man's way of calling this thing; the moment of force is the more quantitative, scientific term which is better at expressing the formula $$ \vec \tau = \vec r \times \vec F $$ The position $\vec r$ and the cross product, for this specific case, are responsible for the words "moment of" while $\vec F$ is the force. A ...


6

The basic answer is that mass scales with the cube of linear dimension and strength of things like legs scales with the square of the linear dimension. Note that large animals have therefore evolved comparatively thicker legs than smaller ones. Linearly scale up a dog to elephant size, and its legs would snap. Even more extreme, think of scaling a ant to ...


5

$$ \text{bhp} \propto \text{torque} \times \text{engine speed} $$ To start with a simple example consider linear motion. If we have some force, $F$, the as the force moves a distance $d$ the work done is force times distance, $Fd$. The power is the work done divided by the time, $Fd/t$, but $d/t$ is just the velocity (distance divided by time) so the power ...


5

Power is force multiplied by velocity. The engine power is actually (relatively) constant regardless of the gear. So when people say there is "more power" in a lower gear, it's the common misconception that "more powerful" is "more forceful" but that's only part of the equation. So if P is constant, then that means if you can combine a large force and a low ...


5

Since $ F = \phi(r) \vec r $, you can find the torque around the origin. Torque $ \tau = \vec F \times \vec r = \phi (r) \vec r \times \vec r$ But $\vec r \times \vec r$ is zero, so the torque around the origin is also zero. Since torque is just rate of change of angular momentum $\frac{ d\vec L}{dt}$, the angular momentum doesn't change, which is what ...


5

The horizantal beam on such scales is intentionally placed below the rotational axis. As long as the weights are in equilibrium the torque is equal on both sides. But as soon as the position changes e.g. tipping the left scale down, the torques differ because only the tangential part of the gravitational force vector in relation to the rotational axis ...


5

OP wrote(v1): So the torque should not be measured in N⋅m but rad⋅N⋅m. Would that then be completely consistent? No, that would not be consistent with the elementary definition of torque $\vec{\tau}=\vec{r} \times \vec{F}$ as a cross-product between a position vector $\vec{r}$ and a force vector $\vec{F}$. An angle in radians is the ratio between the ...


5

It is hard to guess without seeing Gorillapod in use, but my guess would be the following: Center of mass could be understood as an average position of the mass of the object. In order for an object to be in stable equilibrium, its center of muss must be vertically above the area, which is enclosed by contact points of tripod's legs with the ground. If ...


5

I'm not sure if this is the dominant factor, but... Once the coin begins to tip at all, there is torque due to gravity. If you work it out with your hands you'll see that this torque acts perpendicular to the angular momentum from the rolling of the coin and in the plane it rolls on. Thus, it acts to direct the coin in a circular path. A quick experiment ...


5

A net force on an object likes to act through the centre of mass of that object; in this case, the middle of the door. Apply a force to the centre of mass and it will accelerate uniformly because there is an even amount of matter on all opposing sides that can resist the motion with an even amount of inertia. If a force is applied off the centre of mass, it ...


4

Yes, torque has units of joules in SI. But it's more accurate and less misleading to call it joules per radian. Let's take the simple case of a single force acting perpendicular to position (reference) vector: $$\tau=rF$$ To tease out energy from this equation, let's consider an infinitesimal change in (rotational) energy $dE$: ...


4

Every electric motor runs using the Lorentz force, so there is no difference in principle between the motor in the hard drive and an electric car motor. There are commercial 60 hp electric motors that run electric cars, so the answer is yes. It's power consumption is, well, 60 hp. That's 45000 watts, plus a little more for heating the engine and so on, so ...


4

Figured it out thanks to Rex Kerr's comment. The dirt the pole is sitting in is providing a counter Force to prevent the pole from moving, resulting in a net Torque of 0.


4

So remember that because torque is the cross product, we know that it is $rF\cos\theta$. So you can use this to calculate how much torque the pole is feeling. So it is r=3m, F = mg(~1000 N). And $\theta$ is 60. So the torque the pole feels is 3*1000*cos(60) ~ 1500 Nm. The pole can withstand 9000 Nm. Therefore it does not bend. So there is 0 net torque! ...



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