# Tag Info

11

anyway, how likely is it the ice ages could be explained by the earth 'realigning' so that polar regions would migrate over the surface of the earth? How about zero? The geological evidence of the Ice Ages clearly says that, between the ice episodes, the ice did not move. It's just that the polar caps shrank. For instance, the extent of the last ice ...

6

Your intuition was correct - the shaft will rotate in one direction and the housing/stator will rotate in the other. If you look up "moment of inertia" you will find that it is the rotational equivalent of mass. For almost any reasonable motor the moment of inertia of the shaft/rotor windings will be smaller than the moment of inertia of the housing/stator. ...

4

If the bearings were to be considered frictionless, then the maximum speed of the fan will not decrease, though it will take the fan longer to reach the maximum speed. Because as the moment of inertia of the impeller increases its angular acc. will decrease (for the same torque applied), therefore it will take the fan longer to reach its maximum speed. The ...

3

This will depend on exactly what kind of motor you have. If your fan is a brushed DC motor, then the fan speed will be slightly lower, since the new impeller is heavier than the old, so there will be slightly greater bearing friction. The added friction will serve as a power loss, and the motor will have to run slightly slower. If the motor is a brushless ...

3

They don't want to get in front of the rolling wheels -- at least, not intrinsically. When you are turning, your front wheels stop going forwards so much, and start going sideways. As Newton taught us, a great way to think about such things is that an object in motion will tend to keep moving however it is moving. There are sideways forces on the front ...

3

The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by: $$D_t = d_t + \omega(t)\times\tag{1}$$ where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body. Since there there are no torques ...

3

Do I need to use the angular velocity vector in the rotating or inertial reference frame for this? Yes. You can do it either way. I start with the expression that relates the time derivative of a vector quantity $\boldsymbol u$ in the inertial and rotating frames: $$\left(\frac {d\boldsymbol u}{dt}\right)_I = \left(\frac {d\boldsymbol u}{dt}\right)_R ... 3 To add to David Hammen's answer on the question: When numerically integrating this, together with Euler's equation of rotation, is there a way to ensure that the determinant of R remains equal to one (otherwise \vec{x}(t) will also be scaled)? Method 1 Dumb But Effective Naïve Multiplication Whilst you are getting up to speed with more ... 2 The basic physics in laymen's terms Okay, so the basic idea is: an object in motion tends to stay moving at the speed that it's moving. When we apply this to rotational dynamics we have an interesting effect: an objects speed goes linearly with the radius it is from the center it rotates around. So if something is rotating with a period T, it must go a ... 2 Notice that you have implicitly chosen to measure angular momentum about the axle of the platform. That means that all the forces exerted by the axle on the platform are applied through the axis for rotation, meaning the torque they exert is$$\text{force} \times \text{lever arm} = F \times 0 = 0\,.$$And there are no other forces present expect those ... 2 The mistake is in the second line, in the calculation of the differential mass element. The differential mass element in this case is a disc, of radius r where r = R \cos\theta as you have correctly used. However, the thickness of this differential disc is NOT  R d\theta but Rd\theta cos\theta. Try to wrap your head around this. Rd\theta is the ... 2 First, braking causes the car to pitch forward on its suspension and decrease the static friction available on the rear tires and increase that on the front. The rear tires begin to slide because the available friction is less than the friction to prevent sliding. This means they are now in the realm of kinetic friction. The kinetic friction on the rear ... 2 I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity \boldsymbol q, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by$$ ...

2

It will spin about its axis. In general terms, if you attach a flywheel to a big motor in space and turn it on, the flywheel will spin in one direction with some angular velocity, and the motor and whatever is attached to the motor will spin in the opposite direction with a different angular velocity. This is the basic principle behind a reaction wheel. ...

2

@ChrisDrost's answer is correct, but we can actually remove the assumption that the friction is constant by considering conservation of angular momentum instead. If we put our origin at a point along the ground, then there is no net torque on the sphere: The frictional force always points directly towards (or away from) the origin, and the normal force ...

2

Since a system must obey the law of momentum conservation, the center of mass of a system (which can be made of one or many bodies) must have constant velocity if no external force is applied. Hence, a body can rotate around its center of mass, or it can rotate around any other point, but only if under the influence of an external force. Therefore one can ...

2

The astronauts working on the Hubble space telescope had to bring special low torque wrenches to counteract the effect of the torque of the motor spinning them around, due to conservation of angular momentum, although this meant far more use of muscular power to hold them in place. And also to avoid damaging the equipment they worked on, such as screws ...

1

This issue is a bit confusing because there are two types of angular momentum. There's spin, where a rigid body rotates about an axis through its center of mass, and there's orbital, where the center of mass of a rigid body rotates about an axis. For example, the Earth spins about its axis and rotates around the Sun. The total angular momentum can always be ...

1

Depends on the nature of the collision. If there is a mechanism that takes energy from the system, i.e. a deformation, than energy is lost. You could think of your example as the center of mass of your rod as a point mass that starts rotating on a massless string once is passes the pivot. As usual, energy and momentum are conserved. You could do the ...

1

There isn't a simple formula for fan noise, but the physics can be worked out from the fundamental equations of fluid mechanics and acoustics. It isn't a simple problem however. The noise created by fans is complex and from several fundamental sources, and its amplitude depends on frequency. Here is an example of a fan noise frequency spectrum for a cooling ...

1

A spinning top is a gyroscope; it doesn't need air. The gyroscopes on the Hubble telescope work just fine up there.

1

Try to think of this problem using a polar coordinate system. $x$ is essentially the radius $r$ or $\rho$, measured from pivotal. $w$ is simply the angular velocity. So the position vector of the object is $x\hat{\vec r}+\theta\hat{\vec \theta}$ So the velocity vector is $\dot x\hat{\vec r}+w\hat{\vec \theta}$ The hatted vectors are unit. So the ...

1

So this is a phenomenon which is known in billiards as "backspin": you hit a ball off-center and it simultaneously has a motion "forwards" but a spin that imparts a force on the ground to send it "backwards". Trick shots where you induce extreme amounts of backspin by hitting the ball almost vertically downwards are known sometimes as "massé shots", if you ...

1

The velocity jacobian is $\vec{\omega}_B = J\, \dot{q}$ with $q=(\phi,\theta,\psi)$. This is used to transform between the generalized forces/torques $Q$ and the vector torques $\vec{M}_B=(\tau_x^B,\tau_y^B,\tau_z^B)$ $$Q = J^\intercal \vec{M}_B$$ The power through the joint is  Q \cdot \dot{q} = Q^\intercal \dot{q} = \left(J^\intercal ...

1

The same angular velocity of the pedal do not means same angular velocity of the wheel. Assume a chairing with radius $r_1$ and angular speed $\omega_1$, and the cassette with angular speed $\omega_2$ and radius $r_2$ (considering the cassette or the wheel do not make any difference). The speed the chain rolls reads: $r_1 \omega_1=v=r_2 \omega_2$. From this ...

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