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1

You are correct that any forces acting on the axis cannot transmit a torque relative to that axis. But the shaft of a fan is a macroscopic object and consists of material that is measurably away from the spin axis. This deviation is sufficient for it to supply a useful torque.


0

In theory there is no torque from friction once the cylinder is rolling. Imagine that there is no supporting surface but no gravity - then the cylinder will just continue to "roll" forever. So all the supporting surface needs to provide to the cylinder is a force through its centre of mass to counteract the gravitational force. This is a force perpendicular ...


0

If the weight if the panel is $W$ and the distance between the pivot axis and the center of mass is $c$ then the torque on the motor is $$ T(\theta,\ddot\theta) = (I+m c^2) \ddot\theta + W c \cos\theta $$ where $m=\frac{W}{g}$ is the mass of the panel, and $I=\frac{m}{12} \ell^2$ is the mass moment of inertia. Here $\ell$ is the height of the panel. NOTE: ...


1

Does a body always rotate purely about its center of mass? Well, that depends. The first assumption you need is that the body is rigid. Violate this assumption and all bets are off the table because you can't even necessarily classify all motions as "rotations": for example if a long thin board starts twisting sinusoidally into/out-of a helix shape, ...


0

A body in free motion does not necessarily rotate about the center of mass. The center of mass might have straight linear motion in addition any rotation. The general motion is a screw motion with a rotation about some instantaneous axis and parallel translation at the same time. Consider an arbitrary body rotating by $\vec{\omega}$ and at some instant the ...


0

Torque = P*E ( P & E are vectors, also * means cross product) Since the torque is maximum P is perpendicular to E so we can forget about the cross product(sin90 = 1) Torque given here = PE In a capacitor E = potentia/distance between the plates Therefore P = max torque/(V/D)


0

As $d\sin\phi$ is maximized when $\phi=\frac{\pi}{2}$ we find that the torque between the two particles is maximized when $\phi=\frac{\pi}{2}$ where: \begin{equation} \begin{aligned} \tau_{\textrm{max}}&=\frac{F_+d}{2}+\frac{F_-d}{2}\\ &=Eqd\\ &=Ep \end{aligned} \end{equation} where in our last step we used the definition of the dipole moment. In ...


0

Here's how I see it: Assume the friction force is proportional to surface area $S$, via (all other things being constant): $F_\text{friction} = \mu S$ with $\mu$ the coefficient of friction. Then for an infinitesimal ring with thickness $dr$ at $r$: $dF_\text{friction} = 2\mu \pi rdr$. And $dT=rdF_\text{friction}=2\mu \pi r^2dr$ Integrating between ...


0

The torque needed at the pivot A to rotationally accelerate an object by $\ddot{\theta}$i s $$ \tau_A = I_A \ddot{\theta} + c_x W $$ where $I_A = I_{com} + m c^2$ is the mass moment of inertia at the pivot $c$ is the total distance between the pivot and the center of mass $c_x$ the horizontal distance between the pivot and the center of mass $W$ is the ...


0

The answer first: given your axis of rotation, which is the x-axis, you are looking for the moment of inertia associated with that axis, which is the first, smallest principal moment ($I_1$). It only becomes that simple because in your setup the coordinate axes and symmetry axes are identical. Now for some comments: As @BowlOfRed also pointed out in the ...


0

It has to do with how the torque is transmitted from the pedal gear and the wheel gear. The chain transmits the force equally, but the change in radius produces a change in torque. So if the wheel and the pedal gears are similar in radius, the behavior is similar to riding a mono-cycle: all torque applied is transmitted unchanged. The opposite case is when ...


-1

I think we should focus on hike of potential energy within us during climbing through a ramp. That`s why, when we climb either by cycle or walking, we feel it bit laborious. So, taking higher gear ratios, will increase the velocity definitely but it will take more effort making it MORE laborious than usual. I am sharing a link of a video tutorial with ...


0

It seems you have no issue with determining angular acceleration in force systems. For 2D: $$\dot \omega = \tau / J = F r / J$$ where $\dot \omega$ is the angular acceleration, $\tau$ is the net torque applied, and $r$ is the distance between the centre of mass and the line of action of the force. As for translational motion (i.e. straight motion), you ...


3

Torque is a mathematical object called a bivector, produced by taking the wedge product of $\mathbf{r}$ and $\mathbf{F}$. Bivectors can be thought of as area elements of planes; the magnitude is equal to $rF \sin \theta$ and the plane itself contains $\mathbf{r}$ and $\mathbf{F}$. By a great coincidence, in three dimensions, there are three distinct planes ...


2

Torque is a vector whose direction is always out of plane. The same with angular velocity $$ \vec{\tau} = \vec{r} \times \vec{F}$$ $$ (0,0,\tau) = (x,y,0) \times (F_x,F_y,0) = (0,0,x F_y - y F_x)$$ $$ \vec{v} = \vec{\omega} \times \vec{r}$$ $$ (v_x,v_y,0) = (0,0,\omega) \times (x,y,0) = (-y\, \omega,x \,\omega,0)$$ I always think of planar quantities as a ...



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