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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

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

No, these building are still tiny compared to earth's crust mass distribution. One would need to build whole mountain ranges to detect changes in earth gravity field with high precision instruments. And even those wouldn't changed earth orbit measurably because even a mountain range is tiny compared to the mass of the whole earth. However mountain ranges ...


5

When you compute the energy of a rotating object you either consider the tangential velocity or the angular velocity. The two expressions are two different ways to look at the same observable, they both lead exactly to the same result and summing them is a mistake. For instance for a point like mass on a straight line we have: $$E_l = \frac{1}{2}m v^2$$ ...


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 ...


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

No, it does not gain energy. The confusion arises because there's a force that does no work. If the car moves a distance $d \vec l = \vec v dt$, then the work done during that time $dt$ is $dW_{rope} = \vec F_{rope}\cdot d \vec l= 0$. This follows because $ \vec F_{rope}$ and $d \vec l$ are always perpendicular to each other (draw and check this!) which ...


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

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

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

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 ...


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

Potential energy is energy of a position or orientation relative to other positions or orientations. For example, if the hamster wheel is on a table, it will have more gravitational potential energy than if it is on the floor upon which the table is standing. If the hamster wheel is rotationally symmetric about the axis of rotation, and the axis is not ...


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 ...


2

The earth rotates on its own axis, revolves around the center of mass between the earth and moon, and revolves around the sun. If the earthed stopped rotating around its axis, there would obviously be no Coriolis effect and the climate would be significantly different. No more Hurricanes or rotating air masses. Only temperature differences would move air ...


2

Static friction force arises whenever there is interaction between two bodies by direct contact (touch). There need not be any mutual motion between the bodies. This friction force is necessary to explain why the bodies around us maintain their position so reliably. Without friction forces, there would be nothing opposing their mutual motion and the world ...


2

The Earth is currently rotating at one revolution per sidereal day. Converting to radians, this is an angular velocity of $\omega_0 = \frac {2\pi}{\text{sidereal day}}$. You want to have 360 solar days per year, or 361 sidereal days per year. That means a rotation rate of one revolution per 1/361 tropical year, or an angular velocity of $\omega_1 = \frac ...


2

KERS is also known as regenerative braking. It is a system found on the latest hybrid cars. When a vehicle is stopping, the brakes are applied to the discs to generate friction to slow it down. This generates heat which is energy that is lost or wasted. Regenerative braking creates friction in addition to the brakes, however, some of the energy is ...


2

KERS systems (as used in F1 and leMans cars) rely on a flywheel to store kinetic energy. When braking, the transmition is clutched trough a 1:n gearbox to the flywheel. That is, one full revolution of the car's wheels will make the flywheel turn n revolutions. The flywheel system will then store the energy until it is needed. Since a flywheel system rotates ...


2

This is not a dumb question: millions if not billions of people have been taught, incorrectly, that a bicycle's stability originates in the gyroscopic angular momentum of the spinning wheels. This is false, and among other things, would make it near-impossible to turn a bike if it were true. In fact, a bike's stability derives from the momentum vector of ...


2

Wow, I was able to find a solution, but it is super complicated. I can offer though that the change in velocity when it hits the next step is $$ \Delta v = \frac{m r^2 (\cos\varphi-1)}{I+m r^2} v_1$$ where $v_1$ is the impact (drop final) speed, $I$ is the mass moment of inertia, $r$ is the sphere radius and $\varphi$ is the sweep angle going from step ...


2

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 ...


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 ...


2

The answer to this question has essentially been given by @DheerajKumar, but let's explain it in a different way. You are studying circular motion, that is the motion of a particle along a circular path. In such motion, linear speed $v$ gives you the length of arc covered by the particle per unit of time. With angular speed $\omega$ you want to give a ...



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