# Tag Info

144

I'll propose a theory, and I'll describe an experiment I did to test it. Both suggest that the "over" configuration is better, at least if the goal is to make the squares easier to rip off with one hand without making the roll spin out of control. Terms of use Please don't use this post as ammunition to defend a preference. As rightly emphasized ...

78

Take a look at this picture of a cup slightly out-of-balance : In case (A), generated torque is directed out of your reference axis and in case (B) - towards your reference axis. So in case A), you need to compensate out of balance movement with your finger contra-movement. But in case B), torque assists you and makes balancing for yourself, so that you ...

18

You are confusing the ideal case with the non-ideal case. Ideal scenario: No elastic deformation (so no internal work done), no air resistance, no uneven normal forces due to a flexible or soft surface that can cause counteracting torques, no kinetic friction forces. In this scenario, nothing does work and no energy is lost. The kinetic energy will remain ...

15

Maybe because when the cup is the right way up, it’s centre-of-mass is above the point on your finger meaning that as your finger tries to balance the cup any small motion will generate a torque about this c.o.m making it harder to balance. When the cup is upside down, you have your finger on or going through the c.o.m and so any small motion by your finger ...

11

A rolling sphere slows down only when you stop applying the torque (basically when you instantaneously push a sphere forward rather than constantly applying the torque). In the former case, once you have applied the instantaneous torque on a sphere, it will experience somewhat of a counter torque as its rolling due to the irregular surface of the sphere. ...

8

is there a sensible way to introduce a point electric dipole with well defined electrical and mechanical properties? Partly. We have a few options: Finite moment of inertia: For finite $d$, the electric dipole moment is $qd$ and the moment of inertia is proportional to $md^2$. We can take $d\to 0$ with both $qd$ and $md^2$ held fixed, but this requires $q\... 7 Consider an infinitesimally small section of the rope, at an angle$\theta$(measured clockwise) from vertical upward direction (see picture at the bottom) Let the tension be T, on one side of this section and (T+dT) on the other side, both acting in opposite directions (exerting opposing torques, counter clockwise and clockwise respectively). Net torque due ... 7 The professor is doing something tricky, which they should have mentioned explicitly. In order to find a total torque, we need to specify what system we're considering. Suppose our system is the pulley. In that case, the external forces are the force from the support, gravity, the normal force from the curved part of the rope, and the friction from the ... 6 Dr jh has explained by taking torque about the center of mass but it could be explained better if we take it about the point of contact of your finger and the cup. When the cup is upright gravity provides a torque by a force acting on its center of mass which is above your finger. Small perturbations will cause rotation about the point of contact and it ... 6 You only need one force, applied off the object's center of mass. Such a force produces rotational motion, as you know, but it must also produce linear motion. You deliver some amount of linear momentum to the object over the time you are applying the force (because that is what forces do). If there is no other force on the object that can take that momentum ... 6 The two types of engine run on very different principles, so it is not surprising that their torque curves are very different too - it would be more surprising if their torque curves were at all similar. An unloaded internal combustion engine has a natural or "idling" RPM which is determined by the force which the ignition of the air/fuel mixture ... 6 Car A will win, if it has the right final drive ratio. Power is (torque x RPM) so an engine with high power but low torque develops its power at high RPM. Once you gear it down appropriately, it will win the race. 5 In a steam engine, the source of the power is external- you can build up a huge head of steam and feed it at maximum pressure into the cylinder when the the engine is stationary. With an internal combustion engine, you can't build up a source of power externally when the motor is not turning- you have to turn the motor to compress the fuel-air mix and ... 5 Some review just to be sure I hit whatever the gap is. Bear with this: You have three friction coefficients: static, kinetic, and rolling Kinetic A sphere can slide if the force required to angularly accelerate it enough to make it roll exceeds$N$times the appropriate coefficient (kinetic or static to start depending on the situation, and then kinetic). ... 4 The above is obviously not true when considering a force couple. Two separated equal and opposite forces have$\sum F = 0$, but a net torque, and thus$ \frac{\rm d}{{\rm d}t} L \neq 0 $. This is because in general $$\sum \vec{F}_i = 0$$ is not the same as $$\sum \vec{r}_i \times \vec{F}_i =0$$ I might be missing something here, but here is the ... 4 This is an interesting problem, and certainly a model for the forces and torques could be generated. More on this in a moment. However, I think you may run into some issues with the "linear difficulty" curve. The model would tell you the forces, but that doesn't directly correspond to "difficulty." The muscular system can have different ... 4 I can't skip the mathematics bit of the puzzle. Rotation and angular momentum are fundamental properties of a "rigid body." This is the superficial answer, but its an important piece of the puzzle. It lets us reframe your question as to why a rigid body is such a good model of real life objects and their behaviors. This is fun because its so ... 4 It is a 1 newton force applied at a radius of 1 meter, or its equivalent. For example two newtons force at a radius of 0.5 meters would also equal a 1 N$\cdot$m torque and so forth. Force times the radius it was applied at equals torque. 4 why does lower gear generate less acceleration/speed of the car compared to higher gear? You are mixing up acceleration and speed. Acceleration and speed are different and behave differently in the lower and higher gears. Lower gears give higher acceleration and lower speed . Higher gears give lower acceleration and higher speed . To get an intuitive ... 4 The external torque for the system consisting of the boy, the ball and the platform is$0$. The torque you mentioned is an internal torque. Notice that the angular momentum conservation equation includes the angular momentum of all the 3 components mentioned earlier. Internal torque cannot change the angular momentum of a system as internal torque is ... 4 Torque requires a force that is applied to a lever arm. The quoted statement is saying that the gyroscope is free to rotate about the axis AB, meaning that no rotational force can be applied to it as long as it is free to rotate. With no applied force, there can be no applied torque. 4 The statement, "Objects don't rotate when they fall" is false. I know this as an experimental fact b/c I worked on SRTM (https://www2.jpl.nasa.gov/srtm/). Consider the object in the figure: We can approximate this as an asymmetric dumbbell of length$L=60\,$m and two masses,$M=2\,$Gg (the space shuttle), and$m=20\,\$Mg (the outboard radar ...

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Any net force that is not directly in line with the COM of an object, that doesn't have a fixed axis, will cause both spatial translation and rotation of the object. For the object to have translation only, the force must be directly in line with the COM. To have rotation only on the object, equal and opposite forces would have to be applied on equally ...

4

it slows down because there is a net torque on it (by normal force and friction acting in opposite directions) The normal force doesn't supply a torque. It's normal. And if the ball is rolling on a flat surface, in the ideal case friction doesn't change the angular momentum, since it is applied on at the point of contact, and the point of contact is not ...

3

The straw has a (surprising) amount of compressive axial strength, but very little strength away from the axis. In other words, it will bend easily if there are sideways forces. If you place it under moderate compression it is okay, but adding a small sideways force to it at the same time will cause it to buckle. You may not have much information about the ...

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