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0

What he means is: Sum of forces is $\sum \limits_i \vec{F}_i = \sum \limits_i [F_x,F_y]$ Sum of torques is $\sum \limits_i \vec{r}_i\times\vec{F}_i = \sum \limits_i (F_y r_x-F_x r_y)$ So it is not a circular argument because you can have net forces zero but net torque not zero (two equal and oppsite forces a distance apart, or a force couple). In fact, ...


2

The definition of being "in balance" for a rigid body is the absent of acceleration: both laterally and angularly. For a rigid body, the condition of to be "in balance" is the sum of all torque has to be zero with respect to any axis, not just to any particular one. If this condition is not held i.e. there exists an axis that the net torque is not zero, ...


0

In electrical engg, itis very clear in that the current carrying conductor and the magnetic field produced by it are mutually perpendicular to each other.In mechanics it is very difficult to assign a physical meaning for directions of angular velocity,angular momentum and torque.


8

Use the spatial inertia to relate linear/angular momentum to changes in linear/angular speed $$\begin{pmatrix} \vec{L}\\ \vec{H}_A \end{pmatrix} = \begin{bmatrix} m {\bf 1}_{3×3} & -m [\vec{c}\times] \\ m [\vec{c}\times] & I_{cm}-m [\vec{c}\times][\vec{c}\times] \end{bmatrix} \begin{pmatrix} \Delta\vec{v}_A\\ \Delta \vec{\omega} \end{pmatrix}$$ ...


0

To begin I first considered that as the pivot point is in the center of both rods, their masses will not contribute to the amount of torque. Incorrect ... Each rod by itself (considered without the extra masses $m$ attached at its ends) does contribute to the combined moment of inertia of the whole system. The problem statement even provides the exact ...


2

If you attached two flywheels through a motor to the disk at the positions you show, and the motors start spinning in the same direction, then conservation of angular momentum tells us that as the flywheels spin clockwise, the disk must (and will) rotate counterclockwise. However - if you attach the motors to an external structure, you are preventing ...


0

By saying point(c) (or any of the other points) applies a torque on the disc, it sounds like point(c) is a small physical body (If point(c) exerts a torque on the disc, then disc must exert a torque on point(c)) I'm going to assume that the mechanism by which point(c) causes a torque on the disc is a motor connecting point(c) and the disc. So, if points (a) ...


0

On a free floating body, if a pure torque is applied (with net zero force) then the body is going to rotate about it's center of mass (see http://physics.stackexchange.com/a/81078/392). This is regardless of where the torque is applied, or how many torques are applied. If the net force is zero then the center of mass will not move. Now if C or A or B are ...


0

(ANSWER ASSUMPTION: In this answer, I have assume that the stick is not hinged to the rotor, but actually properly fixed. Also, I've interpreted $\tau_B$ as an internal torque/moment in the rotor-stick system because, unless there is some external effect that is causing a torque to be applied, like someone twisting with their fingers at B, there is nothing ...


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The increase in rpm depends upon angular acceleration which is how quickly it's rpm increases (in this case) or decreases. When u increased the rpm, the change appeared instantaneous because of very small time required to increase the rpm. The less the time required, the greater tbe acceleration. So it is a period of acceleration.


1

There absolutely is a period of acceleration. Speed never changes instantly, even if it changes too quickly for you to sense with your eyes and ears, as a direct consequence of Newton's laws. Probably it accelerates over a 1/10 of a second or so, if I had to guess.


2

You forgot to multiply $T \sin{\theta}$ by the distance from the wall to the end of the bar in the torque balance. When you do that, you get an extra factor of 4 in the first term for the expression for x, $x = \frac{8\sin{\theta}}{\mu_s \cos{\theta} + \sin{\theta}} - 2$, which is positive. (PS: I didn't check your math, I just added the factor of 4, so I ...


-4

Because of hp and torque. When in a low gear, the rpms are higher and the hp and torque kicks in at higher rpms.



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