# Tag Info

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 ...

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 ...

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 proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and ...

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

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

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.

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

I'm outlining this and stating the final result so that the OP gets the fun of figuring this out themselves. Future responders, please don't work this out All you have to do is allow $\omega(t)$ to be a function of time. You'll get extra ${\dot \omega} = \alpha$ terms in your equation, and you'll get a final result that says that $${\vec a} = {\vec ... 3 The only way to purely rotate a rigid body about its center of mass is to apply a pure torque (no net force). If the net force applied is zero then the center of mass is not accelerating. However and combination of translation and rotation of the center of mass can be viewed as a pure rotation about the instant center of rotation. So to effectively answer ... 3 In the reference frame of the car, the axle is stationary, but the ground is moving below at speed v_C. If the car doesn't skid, then the surface of the tyre must move at the same speed, but with a velocity that is directed backwards in the bottom and forwards in the top. At half the distance between the tyre surface and the hub, the speed is v_C/2. ... 3 The wikipedia link does an adequate job of explaining this. You can find a very mathematical discussion in this partial duplicate question, but here is my simple take on it. The angular displacement in three dimensions does have a vector nature in the sense of having both a magnitude (the angle through which you turn something) and a direction (the axis ... 3 Let the cone lie on the \hat{X}\wedge \hat{Y} plane (z=0) and let the z axis pierce this plane at the cone's apex. If the cone's half angle is \alpha, then its axis of symmetry as a function of time is defined by the vector$$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 = ... 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. 2 If two eggs are given out of which one is hard boiled egg and the other is raw egg. When both of them are allowed to spin on a tabletop, the egg which spins slower must be the raw egg because the liquid inside it tries to get away from the axis of rotation increasing the value of moment of inertia (I). To remain the angular momentum L conserved in absence of ... 2 The tension in the rope should be different on the left than on the right - it is this difference that gives rise to the torque that accelerates the pulley. You seem to think that it should be the same: but if it was, then where would the torque to move the pulley come from? Annotate your diagram carefully: you did not show T anywhere. 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 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$$where r is the 'radial' distance from the centre of the wheel to the top of the water. In practice, there will be some water sliding past the wheel, depending upon the hydrodynamics of the ... 2 In short, it continues to rotate simply because it has no reason to stop rotating. Imagine I have a puck and I whack it across an ice-rink. The puck will continue to travel at the same speed until it hits a wall; i.e it will stay the same unless there is a reason not to. This is conservation of linear momentum as shown here. The same thing happens with ... 2 In spherical coordinates the position is$$ {\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 ...

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