# Tag Info

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 ...

22

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 ...

15

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 ...

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 ...

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 ...

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 $E=\frac ... 5 What I don't understand is why we have to apply the rule in this specific way? How do I know how the chain rule must be applied? We don't have to. You don't know. Somebody just found out that by using that specific method, the result ended up neat and simple. Nothing is wrong with another method. You get the same thing in another expression. Let's ... 4 Imagine a object steadily traversing a circle of radius$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 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 ... 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 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 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 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 If you start with a monatomic gas then the only degrees of freedom available are the three translational degrees of freedom. Each of them absorbs$\tfrac{1}{2}kT$of energy, so the specific heat (at constant volume) is$\tfrac{3}{2}k$per atom or$\tfrac{3}{2}R$per mole. If you move to a diatomic molecule there are two rotational modes as well - only two ... 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 ... 2 "Rest, in physics, refers to an object being stationary relative to a particular frame of reference or another object." - Wikipedia (emphasis mine) While on Earth, the planet is often treated as the default frame of reference. It is not a perfect frame of reference, but for many purposes it is good enough. Since there is no absolute frame of reference, ... 2 Which is right? So let's first establish what is correct. Suppose we have a circle of radius$r$and mass$m$centered on a circle of radius$\bar R$(not quite what you have defined; in your case$\bar R = R + r$). We'll say that the position of the$r$-circle on the$\bar R$-circle is angle$\phi$and, from the angle$\phi$, we will measure an angle ... 2 You have not considered the direction of torque in you equation. Since the Torque is caused by Frictional Force$F_r$which is in the direction$-i$. the torque is$\vec{\tau}= \vec{r}\times\vec{F_r}$and$\vec{r} $is in$j$direction So the cross product yields$\vec{\tau}= -|r||F_r|\hat{k}$which ofcourse is in the negative direction of positive z-axis 2 Friction is required for a spinning ring or cylinder to simulate gravity. The method works because the inner surface provides a tangential acceleration to objects that contact it, and then a radial (normal) acceleration to those objects which otherwise at this point would travel in a straight line tangent to the rotation. If the wall of the ring were ... 2 The rotational kinetic energy of a (uniform) solid sphere rotating about an axis passing through the center of mass is given by$\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 ...

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