# Tag Info

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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, ...

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To see why the first configuration is used rather than the second, perform the following experiment: Hold a bowl in your hand and place a small ball inside. Move the bowl in circles at various speeds and observe the behavior of the ball. Now turn the bowl over and balance the ball on top. Again, move the bowl around and observe the ball. Which is more ...

11

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 ...

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I can see a difference by thinking about the flanges... Given that 1) the flanges are on the inside of the wheels 2) the right handside of the diagrams is the outer part of the track where the flange will press against the rail... compare and where the red lines indicate the plane of the flange.... in the upper case the flange neatly pushes ...

0

In general, the direction of a friction force is determined by which object you are analyzing, and you only have 1 object in your free-body diagram. The friction will be opposite the direction the object is or is attempting to slide across a surface. It is not necessarily opposite the direction the object is moving with respect to your coordinate system. ...

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Yes, your equation is correct and you can calculate the torque at the engine by measuring the torque at the wheels. However the number you get will be less than the true engine torque because of friction in the drive train. The figure you calculate will be the engine torque minus the transmission losses. Rolling roads estimate the transmission losses by ...

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

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. ...

0

Let's assume that the rotating mirror was at 45 degrees to the emitted beam when the beam reaches it, and this reflection point is the center of rotation. The beam is then reflected by the stationary mirror back along its path to the rotating mirror, so that the return beam intersects the rotating mirror at the same point as it did the first time through. In ...

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Your integrals are correct; I think it's just an integration error: $$\frac{m}{L/2} \int_0^{L/2} l^2 \, dl = \frac{m}{L/2} \left[ \frac{l^3}{3} \right]_0^{L/2} = \frac{m}{L/2} \left[ \frac{L^3}{24} \right] = \frac{mL^2}{12}.$$

0

In order to understand this motion, let us check first the case of a particle's motion in a vertical circle; here the particle of mass $m$ is attached to an inextensible string of length $R$ which provides the necessary centripetal force along with gravity. Let the initial velocity be $u$ at the lowest point of the vertical circle. After time $dt$, it ...

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min. speed will be square root of the product of length and gravity. as per the requirement of centerpital force to complete circular path.

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The angular velocity is not well defined for the center of a wheel, or for the center of mass of a rigid body, in the same sense that the number $0/0$ is indeterminate. This is a consequence of the fact that the center of mass does not move if the motion is purely rotational. To see this, let us start ab ovo... The angular velocity is defined as The ...

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You know the law of motion that states that the net forces applied on a body affect the linear motion of the center of mass only. Complementary to this is the law that the net torques on body about the center of mass affects the rotational motion of the body. Rotation is common to all points on a body so if any part rotates so does the center of mass. To ...

0

Just to clarify there are two different cases:- 1) Purely rotational motion of a wheel where there is no translation of the centre of mass, here assumed to be the centre of the wheel. In this case the centre remains fixed and doesn't rotate. Angular velocity is defined for the entire rigid body as a whole, so it is the same for the centre as for any other ...

1

$$\mathbf{a}=\dfrac{\Delta \mathbf{v}}{\Delta t} \tag{01}$$ \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

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 ...

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 ...

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 $E=\frac ... 3 I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ... 0 First Thought (probably not the fastest) Let us assume you have a vector space in$R^{3}$with a quaternion defined as: $$\mathbf{q} = q_{F}^{*} \ q_{B} \\ = a + b \hat{\mathbf{x}} + c \hat{\mathbf{y}} + d \hat{\mathbf{z}}$$ where$(a, b, c, d)$are the Euler parameters and$(\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}})\$ defines the reference ...

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