# Tag Info

24

It depends what you mean by day and night. The day and night are not of equal lengths now, where I live at latitude 53N. The tilt of the Earth's rotation axis with respect to the ecliptic plane means that this is generally true. The situation you describe would have to be considerably more extreme. If the planet was in a highly eccentric orbit and had a ...

24

I know that the opposite could happen. There is an old book called "Night Fall" about a planet that had three stars. Because there was always a star shinning on all sides it would never get dark. About every 500 years everything lines up just right so that all the stars were on one side. Then as the planet rotated on its normal spin, night fall would come ...

23

Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right. Shift the ...

23

"A state of rest" is a relative term. Relative means - measured in comparison to the things around it. When you sit in a train and sip from a cup of coffee, you can do so because the cup is still relative to you even though both of you might be hurtling through the countryside at 200 km/h. For most experiments, objects can be considered "at rest" if they ...

16

In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, ...

11

What does this small change means in form of Rotational Kinetic Energy? There's a problem with your calculation: You assumed a constant value for the Earth's moment of inertia. The Moon and Sun raise tides on the Earth itself. These Earth tides result in subtle changes in the Earth's moment of inertia. The signature of these tides can easily be seen in ...

7

Imagine a object steadily traversing a circle of radius $r$ centered on the origin. It's position can be represented by a vector of constant length that changes angle. The total distance covered in one cycle is $2\pi r$. This is also the accumulated amount by which position has changed.. Now consider the velocity vector of this object: it can also be ...

7

If planet is going around some star, at least somewhere at the surface of this planet is going to be a day, or else some other celestial body is covering the star light. If the speed of rotation of the planet is arranged in such a way that while it's traveling around the star one side of planet is always facing the star, and other side is always facing the ...

6

The contact with the rail creates a kinematic center of rotation where the reaction forces meet. The rail car will tend to rotate about this center as a result of side loads. If the center is above the center of mass, the rail car acts like a hanging pendulum. A small deflection will cause a restoring torque opposing the swing. If the center is below the ...

6

Revolving around the sun is equivalent to free fall around the sun, so the revolution allows you not to 'feel' the sun's gravity. The rotation of the earth is something that can be measured: (i) a centrifugal force which is a small offset on gravity, and (ii) causes the coriolis force. Both these are small effects, so can often be ignored for laboratory ...

6

One scenario where a planet could have a much longer night than day would be the case if it were in a large elliptical orbit with a large companion, and were additionally tidally-locked to that companion. If the planet in question were to advance its periapsis in lockstep with its year, then the planet could have a long period of being distant from its ...

5

Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac ... 5 What I don't understand is why we have to apply the rule in this specific way? How do I know how the chain rule must be applied? We don't have to. You don't know. Somebody just found out that by using that specific method, the result ended up neat and simple. Nothing is wrong with another method. You get the same thing in another expression. Let's ... 5 When you are walking on a train, why does it not take you longer to walk from front to back than the reverse? Why does it not take more effort? It's because you "share" the speed with the train. At that moment, your reference frame is the train, and only your movement relative to the train affects you. The same applies when you are walking on the earth. Only ... 4 Take the definition of day to be sunrise to sunset, that is all the time that any part of the star is visible from the point on the planet. Given that the diameter of the star is greater than the planet, there is slightly more day than night. (Draw the two lines that are tangent to 2 circles, a larger circle and a smaller circle exterior to each other.) ... 4 As the object spins faster, you will need a higher drive frequency to continue to apply torque. At the size of a baseball, circumference is about 9.25 inches (let's say 25 cm). To rotate the surface at the speed of light would imply 1.2 GHz . At that frequency, the skin depth for steel is about 1.2 um, implying very high resistivity which would lead to high ... 4 Constant acceleration - whichever type of acceleration - means that the second derivative is constant. In the linear motion, you have that the second derivative of position is constant. In rotational motion, you have that the second derivative of the angle is constant. In both cases you have $$f''(t)=c \tag{1}$$ where$f$may be$r$or$\theta$, and$c$... 3 Moments of inertia are additive. Suppose you have particle$A$with a moment of inertia$I_A$and particle$B$with a moment of inertia$I_B$. Then the total moment of inertia of both particles is just$I_A + I_B$. You can imagine a ring as being made up from lots of point particles, all at a distance$R$from the central axis. In that case the moment of ... 3 I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ... 3 When calculating gravitational potential energy the only thing that matters is the position of the centre of gravity. So if the vertical position of the centre of gravity changes by$\Delta h$the gravitational potential energy changes by$\Delta V = mg\Delta h$. 3 Rigid bodies with three distinct moments of inertia have two stable rotation axes, the axes with the greatest and least moments of inertia (typically the shortest and longest axes). Non-rigid bodies have but one stable rotation axis, the axis with the greatest moment of inertia. The axis with the least moment of inertia becomes unstable thanks to entropy. ... 3 I don't think this is as dumb a question as everyone downvoting you seems to think. The definitions of angular velocity ($\omega$), angular momentum ($L$), and moment of inertia ($I$) ARE defined in order to perfectly mirror Newton's laws. Angular momentum is the analogue of momentum, angular velocity is the analogue of velocity, and moment of inertia is ... 3 When initially you exert a force$F_i$to get things going, you're actually exerting a torque$T$about the centre point of the circle: $$T=F_i R,$$ with$R$the radius of the circle. According to Newtonian physics, this torque causes an angular acceleration$\dot{\omega}$as follows: $$F_i R=I\dot{\omega},$$ where$I$is the Moment of Inertia of the ... 3 The essence of the question is the definition of the angular velocity and its relation to the axis of rotation. First of all, to describe the rotation of a single particle, one requires an axis of rotation$\hat{\mathbf{n}}$, and a rate of change of ‘orientation’ (or ‘angular position’)$\frac{d \theta( \hat{\mathbf{n}}) }{d t}$around that axis. Without ... 3 For a smaller object you can do a bit better "Dr Yoshihiki Arita, Dr Michael Mazilu and Professor Kishan Dholakia of the School of Physics and Astronomy at the University of St Andrews were able to levitate and spin a microscopic sphere, purely using laser light in a vacuum, briefly up to 600 million RPM before it broke apart." ... 3 I am not sure what you actually want to ask. So I would recommend that you put more effort into your question. Assuming that this is the setup that you have, a scissor in gray and the cross section of the rope in orange: Say the toque applied to the joint of the scissors is$\tau$. What is this force$F$, then? The distance of the rope contact point to ... 3 It would be easier to answer your question clearly with a drawing. In the following, the angle coordinate of the pendulum is the angle it makes with the vertical line. When the pendulum swings right(left), the angle will be positive(negative). With this setting, I get the exact same answer as you by working out the equations of motion. However, there ... 2 The best way to describe a rotating reference system is via a simple change of coordinates. If you have a description of a phenomenon in some set of inertial coordinates$\{t, x, y, z \}$, then you can obtain a description of its motion in a rotating reference frame with coordinates$\{t', x', y', z'\} $by making an appropriate substitution. For example, ... 2 The answer by Bryson S. is solid, thorough and vey good, as is Jerry Schirmer's hint. This is merely another way of looking at the problem. We can consider, as Jerry Schirmer points out, two components of acceleration; a tangential and a normal component. Before we begin, note that velocity always points tangential to the path a particle travels. This is ... 2 Equation 24 of your paper says: $$\vec{u}=\vec{U}+\vec{\omega}\times\vec{R}$$ Here$\vec{U}$is the position of the center of mass,$\vec{\omega}$is the angular velocity vector, and$\vec{R}\$ is the position of one point in the rigid body. Taking the derivative, ...

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