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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 ...


13

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, ...


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 ...


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 ...


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

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

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 $\Delta V = mg\Delta h$.


2

Firstly, you must qualify moment of inertia by the axis it is taken about. If you translate this point, the inertia also changes as described by the parallel axis theorem. So you question relates to the moment of inertia of a ring about the ring's axis of symmetry normal to the ring's plane, and to a point on this plane at the same distance from this ...


1

In order to move through a concave path, an agent has to impart force to otherwise a linearly-moving object. The object , by virtue of its motion, under the absence of any external force, always travels or tends to travel in the direction of the velocity vector at the concerned instant. So, when the object has to transverse a curve trajectory, the main ...


1

\begin{equation} \mathbf{a}=\dfrac{\Delta \mathbf{v}}{\Delta t} \tag{01} \end{equation} \begin{equation} \vert\mathbf{a}\vert=\dfrac{\vert \Delta \mathbf{v} \vert}{\vert\Delta t\vert}=\dfrac{\upsilon \vert\Delta \phi\vert}{\vert\Delta t\vert}=\dfrac{\upsilon \vert\Delta s\vert}{r \vert\Delta t\vert}=\dfrac{\upsilon \cdot \upsilon \vert\Delta t\vert}{r ...


1

The same angular velocity of the pedal do not means same angular velocity of the wheel. Assume a chairing with radius $r_1$ and angular speed $\omega_1$, and the cassette with angular speed $\omega_2$ and radius $ r_2$ (considering the cassette or the wheel do not make any difference). The speed the chain rolls reads: $r_1 \omega_1=v=r_2 \omega_2$. From this ...


1

Moment of inertia depends on how far away mass is from the axis. In a ring of radius $R$, all the mass is $R$ from the axis. For a single particle $R$ away from the axis... well, all the mass is $R$ from the axis.


1

For pure rolling (no slipping) the ball is rotating about point B (the contact point). Thus A is moving up and a little to the right, and C downwards and a little to the left, whilest point B is at rest. If B was moving up or down it would break the contact, and if it move left or right it would slip. Why? The picture below shows the velocity vectors ...


1

There are 2 kinds of motion here: translational and rotational. The net velocity at any point is calculated by the vector sum of these two velocities. For instance at point B, $v-rw=0$ (pure rolling condition). That's why this point stays at rest.



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