# Tag Info

1

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 ...

0

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.

2

Yes. The solution is: $$\bf{r} = \dfrac{\left( \sum {\bf F}_i\right) \times \left( \sum ({\bf r}_i \times {\bf F}_i) \right)} {\| \sum {\bf F}_i \|^2} =\dfrac{{\bf F} \times {\bf \tau}}{{\bf F}\cdot{\bf F}}$$ Then you can show that $${\bf r}\times \left(\sum {\bf F}_i \right)= \sum ({\bf r}_i \times {\bf F}_i) = {\bf }\tau$$ Use ${\bf F} =\sum {\bf ... 0 Let me introduce the notation $$\sum F_{i,x} = F_x, \ \ \ \sum F_{i,y} = F_y, \ \ \ \sum F_{i,z} = F_z, \tag{i}$$ Since the determinant is zero, there may be indeed, no solution of the system. But if the system of equations has a solution, recall that the body doesn't rotate around a point, but around an axis. So, your$\vec r\$ is bound to be on an ...

0

Assuming that this is a competition of some sort, and everyone has the same set of powertrain components (motor, batteries), then the best way to maximize the output of that powertrain is to characterize them and find the peak output in terms of torque at a given speed. It would be valuable to find this at a range of speeds starting from 0 so you can ...

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