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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.


8

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 ...


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 ...


6

The inertia ellipsoid is computed from an integral about an axis - in other words you rotate the object. This will "smooth out" any symmetries and typically increase the symmetry. Sorry this is a "early morning" intuitive explanation - maybe someone else will give you a more formal answer.


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 ...


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

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

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 ...


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.


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

Angular displacement is an example of what's generally called a pseudovector. This is a quantity that is similar to a regular vector, except for the fact that it behaves differently under improper rotations such as reflections (it gains an additional sign flip). Any quantity which is the cross-product of two polar vectors will generally be a pseudovector. ...


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 ...


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

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

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

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

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 ...


2

There is no such thing as "rotational speed". There is angular momentum, which has units of $\frac{\mathrm{kg} \cdot \mathrm{m}}{\mathrm{s}}$ (and so is not a speed), and angular velocity which has units of $\frac{1}{\mathrm s}$ (and so is not a speed either). There is no maximum angular velocity in special relativity. So long as $\omega r\lt c$ (the ...


2

The rotational kinetic energy of a (uniform) solid sphere rotating about an axis passing through the center of mass is given by $\frac{1}{2}I\omega^{2}$, where $I=\frac{2}{5}MR^{2}$. So $K=\frac{1}{5}MR^{2}\omega^{2}$. Using $M=6\times10^{24}\,\mbox{kg}$, $R=6400\,\mbox{km}$, and $\omega=\frac{2\pi}{T}$, with $T=24\,\mbox{hrs}$, we get ...


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

Physics does not have a proper, rigorous concept of causation. There are the terms locality and causality, but they are technical terms with precise meanings that do not occur in Newtonian physics. Nevertheless, Newton's law, $\vec F = m \ddot{\vec x}$ is often seen to embody causation in a certain sense: You are given, as external circumstance, the total ...


2

When you fix a reference point (take it to be the origin of your reference frame) you can write the position as $$\vec{r} = r \hat{r}$$ where $\hat{r}$ is the unit vector pointing toward the particle. Deriving you obtain $$\vec{v} = \frac{dr}{dt} \hat{r}+ r \frac{d \hat{r}}{dt}$$ The first term is the radial component of the velocity, the second one is ...


2

In spherical coordinates the position is $$ {\bf r} = r \ \hat{r} $$ and the velocity is (see this) $$ {\bf v} = \dot{r} \ \hat{r} + r \dot{\theta} \ \hat{\theta} + r \sin \theta \dot{\phi} \ {\hat{\phi}} $$ The angular momentum is then $$ \begin{eqnarray} {\bf L} = {\bf r} \times m {\bf v} &=& m r \ \hat{r} \times \left(\dot{r} \ \hat{r} + r ...


2

Assuming the wheel moves at the same speed as the water (ie: neglecting and 'slip' of the wheel past the water), angular velocity is given by: $$\omega = v / r$$ where $r$ is the 'radial' distance from the centre of the wheel to the top of the water. In practice, there will be some water sliding past the wheel, depending upon the hydrodynamics of the ...


2

In short, it continues to rotate simply because it has no reason to stop rotating. Imagine I have a puck and I whack it across an ice-rink. The puck will continue to travel at the same speed until it hits a wall; i.e it will stay the same unless there is a reason not to. This is conservation of linear momentum as shown here. The same thing happens with ...


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 ...



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