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-1

Here is my answer. I think it's self explanatory enough: In the last sentence there must be a minor correction: If someone keeps omega1 constant, then for the same speed they have to cycle faster so they can maintain the same power. If someone keeps omega2 constant, then bigger alpha mean less forward speed but less power too.


1

The power input is roughly constant (that of a car is dictated by the total engine power while for a bicycle it depends on the user). The gear or similar tools adjusts the mechanical advantage so that a low gear will express the engine power in force rather than speed (recall that power is force times speed). On higher gears the force is traded in for speed. ...


0

There are cases where you can do as you suggested and where you can't: A: If the external forces exerted on the rotating shaft doesn't change as the shaft accelerates, then you can simply apply the formula you suggested: Measure the MoI once, and use it at all times. Since the external forces are constant (or zero), your MoI will not change. Example: You ...


1

You can. The problem is practicality, not physics. When you use a dynamometer, you can calibrate the meter once and use it for all the measurements you want. When you use the MOI you have to measure/calculate it for each measurement, as it is part of the unit under test. Sometimes you don't have access to the necessary data. Also measuring under ...


1

The center of mass is defined as it is because the center of mass obeys Newton's second law. Consider a collection of particles, labeled by an index $i$. Then each particle obeys $$\mathbf{F}_i=\dot{\mathbf{p}}_i$$ We decompose the force into the interaction between particles and an external force $$\mathbf{F}_i=\sum_{j\ne ...


2

Imagine a thin stick that you place on the floor and you want it stand. If you succeed to put the stick so as its weight, considered as acting on the center of mass, pass through the point $O$ of contact with the floor, the stick will stand. Otherwise it will fall. Let's see why. Let's decompose the weight of the stick into a component along the stick, and ...


0

Three parameters are needed for 2D force (as opposed to 6 for 3D, see http://math.stackexchange.com/a/1157906/3301). Composition A force with magnitude $F$ along a direction $\vec{e}=(e_x,e_y)$ going through a point $\vec{r} = (r_x,r_y)$ is described by the three parameters $$ f =(a,b,c)= ( F e_x , F e_y , F (e_y r_x - e_x r_y) ) $$ Decomposition Given ...


1

Yes your intuition is correct. Absent the weight of the pipe and your ratchet, you will produce 200 ft-lbs this way. I suspect the weight of the pipe is a small error compared to others in the system, but if we model the pipe as a uniform beam we can check. Let the mass of the pipe be $m$ lbs. The mass of a small length $dx$ is then $\frac m4 \ dx$ lbs. ...


2

Looking at your video, it appears that the boomerang is not turning fast enough to return. This typically means that it is too heavy. I wrote a couple of answers earlier [here](http://physics.stackexchange.com/a/156122/26969_ and here to explain some of the physics of boomerangs; perhaps you will find the physics there hard to understand, but then these are ...


2

The change in angular momentum of an object is equal to the integral of torque over time: $$\Delta \vec{L} = \int \vec{\Gamma}dt$$ Note that I wrote both angular momentum and torque as vectors. When torque is along the axis of rotation, you get either an acceleration or deceleration of the angular motion (depending on the relative direction of each). When ...


1

What you have to assume a defined motion, where $\theta(t)$ is known. If the mass was spinning with a constant rate there would be no torque needed. The basic equation of motion is $$T = I_{zz} \ddot{\theta}$$ where $I_{zz}=\int r^2 {\rm d}m$ is the mass moment of inertia about z. That internal torque is the translated into shear stress with $$\tau = ...


5

John Rennie's answer is correct in the special case that angular momentum and angular velocity are parallel to one another. This is not always the case as moment of inertia is a second order tensor. Angular momentum is given by $\boldsymbol L = \boldsymbol{\mathsf{I}} \boldsymbol \omega$. Differentiating this gives the rotational analog of Newton's second ...


2

The equivalent to force is torque, $\tau$. So the second law would be: $$ \tau = I\dot{\omega} = I\ddot{\theta} $$


1

Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that $\mathbf M_O$, $\vec{OP}$ and $\mathbf F$ belong to $\mathbb{R}^3$, as MyUserIsThis did in his comment. If this is not satisfactory to you, you could consider three distinct fields of numbers, say $\mathbb{R}_F$ for forces, ...


4

The torque and power curves are of course related (power is related to torque times RPM) but the maximum torque and maximum power are achieved at different RPMs and tell us different things about the car, therefore both are specified. Best is of course to look at a diagram like the one below, showing the relationships of the BMW 335i twin-turbo engine as ...


1

Power will tell you how fast you can go - without power you cannot overcome the drag (force of drag goes with $v^2$ so power needed goes as $v^3$). The torque is a measure of the instantaneous acceleration you can get - the ability to transfer power to the wheels. Note that acceleration also requires power - you are adding kinetic energy - but at lower ...


3

Torque ($\tau$) and power ($P$) are mathematically related to each other: $$P = \tau \omega$$ Where $\omega$ is the angular speed (effectively the RPM) of the engine. Since the torque of an engine is a function of $\omega$, we can think of this as either $$P(\omega) = \tau(\omega) \omega$$ Where $\tau(\omega)$ is some torque-speed curve dependent on the ...



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