# Tag Info

21

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 ...

12

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, ...

9

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ...

8

Radian $(\theta)$ is defined as, $\theta=\dfrac{l}{r}$, where $l$ is length of arc and $r$ is radius in a circle, and both have dimension as lengths. Thus, Radian is a dimensionless unit.

6

I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become $$F\delta t$$ (from ...

5

Yes, $B$ does rotate when seen from a static frame of coordinates outside the disk: As to velocities and accelerations, see the article in Wikipedia. It says, $$\vec {v_s} = \vec {v_r} + \vec {\Omega} \times \vec r,$$ where $v_s$ is the velocity in the static frame and $v_r$ in the rotating. If you apply this formula for both points $A$ and $B$, their ...

5

If I understand your question correctly you are saying that: $$v = r\omega$$ and therefore: \begin{align} v_A &= r\omega \\ v_B &= \tfrac{1}{2}r\omega \\ v_A &= 2v_B \end{align} but how can $A$ and $B$ have different velocities when they are both attached to the disk so the separation between is fixed? The answer is that $A$ and ...

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 $E=\frac ... 5 A day is currently about 86400.002 seconds long. If we could just increase the Earth's rotation rate by a mere 2 milliseconds per day we would get rid of the need for those pesky leap seconds. No problem! We only need something that rotates with an angular momentum of 1.4×1026 joule-seconds about an axis pointing due south. One way to do this would be to ... 5 The key is the coriolis force. The coriolis force is$F_c = -2m\Omega \times v $. Here$\Omega$is the rotation of the frame of reference and$v$is the linear speed of the satellite. If you do the calculations, left as an exercies for the reader, you'll get the missing force. In case 2 the coriolis force is 0, because the velocity$v$has to be used in ... 4 I have reproduced your calculation. If$\theta$is the angle from the central vertical axis of the hemisphere to the ball, the tangential downward force is$g \sin \theta = g\frac rR$The tangential upward force due to rotation is$\omega ^2 r \cos \theta=\omega ^2 r \sqrt {1-\frac {r^2}{R^2}}$When$\omega \lt \sqrt{\frac gR}$the downward force is ... 4 Imagine a object steadily traversing a circle of radius$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 ... 4 The point of contact of the wheel with the street is actually stationary, provided the wheel does not slip. This is due to assuming the friction the street exerts on the wheel at the point of contact is sufficiently high. However, this is only true for a particular instance! The very next moment, the point of contact, P, is no longer is contact with the ... 4 If no force acts on a body it moves in a straight line. To make the body deviate from a straight line you have to apply a force to it. Therefore applying the centripetal force to the body is what makes it move in a circle. If you apply a constant force at right angles to the direction of motion then your object will indeed move in a circle. 3 I'll make an example, to make things clear. Take a two body system, in which the particles are seperated by a constant distance$d$and have mass$m_1 = m_2 = m$. This is a holonomic constraint, since $$| \vec{r}_1 - \vec{r}_2 | = d$$ with the particle-positions$\vec{r}_1$and$\vec{r}_2$. This system is therefore reduced to 5 degrees of freedom (6 minus ... 3 In the reference frame of the car, the axle is stationary, but the ground is moving below at speed$v_C$. If the car doesn't skid, then the surface of the tyre must move at the same speed, but with a velocity that is directed backwards in the bottom and forwards in the top. At half the distance between the tyre surface and the hub, the speed is$v_C/2$. ... 3 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 ... 3 The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ... 3 How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force? If you do the math (equating$G M m / R^2$to$m \omega^2 R$and using$M = \frac{4\pi}{3} \rho R^3$as well as$\omega = 2\pi f$), it follows that the size of the sphere is entirely irrelevant and that only the ... 3 Let the cone lie on the$\hat{X}\wedge \hat{Y}$plane (z=0) and let the$z$axis pierce this plane at the cone's apex. If the cone's half angle is$\alpha$, then its axis of symmetry as a function of time is defined by the vector $$A(t)=\cos\alpha \left(\cos(\omega_0\,t) \hat{X} + \sin(\omega_0\,t) \hat{Y}\right)+\sin\alpha \hat{Z}$$ where$\omega_0 = ...

3

The wikipedia link does an adequate job of explaining this. You can find a very mathematical discussion in this partial duplicate question, but here is my simple take on it. The angular displacement in three dimensions does have a vector nature in the sense of having both a magnitude (the angle through which you turn something) and a direction (the axis ...

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

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

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 ...

2

Angular displacement is define by change in angle i.e. ${\Delta \theta}=({\theta}_1 -{\theta}_2)$, where $\theta$ is taken as positive in anti-clock wise. Hence angular displacement ${\Delta \theta}$ has both magnitude and direction. But let when a particle rotate from a point A in $\theta$ angle in anti-clock wise to point B and then back through $\theta$, ...

2

You've almost got it. You would want to express everything in terms of just one variable, otherwise you run into trouble. Right now you have both $\omega$ and $\theta$, but there is a relationship between them... However, since you are asked for an expression in $\theta$, it is simpler to use conservation of energy - you don't actually need to know the ...

2

Yes, this is certainly true. Mass is defined by $m^2=E^2-p^2$ (in units with $c=1$), where $(E,p)$ is the momentum four-vector built out of the mass-energy and momentum. (This defines what's known as invariant mass, as opposed to "relativistic mass.") Mass as defined in this way is not additive, and depends on the motion of the particles within a system. As ...

2

what does the commentator mean by concavity of the floor? He or she means that the surface you walk on is in fact the inside of a cylindrical surface. Like a very large version of the inside of a wedding-ring. The curvature of this surface can be measured.

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