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The yaw angle is the rotation around a vertical axis. It is easiest to understand if $\theta_l = 0$. In that case, the l wheel stays fixed and the r wheel travels in a circle around it. The distance the r wheel travels is $R*\theta_r$. The robot rotates through an angle $\phi = R/W*\theta_r$. If $\theta_r > \theta_l > 0$, the robot drives around in ...

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You can think of $R*\theta r$ and $R*\theta l$ for small $\theta$ as the linear displacements of the right and left wheel contact points with the ground and, without the wheels slipping, the difference causes a 'twisting' of the body over the wheel separation, $W$ in the vertical axis (yaw) as the wheels independently turn one direction or another. The ...

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I'm not knowledgeable in some aspects of the question, but I will provide an answer unrelated to others. An object can rotate so fast that some representations of angular velocity cannot be valid. For example, when an object rotates more than 180-degrees or 360-degrees (pi or 2*pi radians) per unit time, the representation must be able to represent such ...

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I've got v(0.4)= 3.066 m/s I'will use physical pendulum, circular geometry - hyperphysics with data : radius 0.15 cm , g=9.8 m/s/s The period T for a sphere is 0.09197 sec (last value in the page) (is more than 10 times faster than a simple pendulum of length 30 cm) I get $$\theta(t)=\theta_{max}*cos((t-t_{0})*2\pi/T)$$ With the substitution of sin ...

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For a complete description, see http://www.algobeautytreb.com/trebmath356.pdf For a FORTRAN listing which includes a bunch of stuff you probably don't need, http://www.algobeautytreb.com/fortmontecarlo.html And, despite what some would have you believe, it is not simple at all. Among other things, neither the arm nor the sling undergo uniform angular ...

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As the car turns around, since the car is a rigid body(ideally), not only will the rear wheel experience a centripetal force, but also other parts of the car. You can illustrate this by recalling your experience in a car when the car turns around. You will feel a force pushing you in the direction of rotation. So every part of the car is experiencing a ...

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When the car takes the curve, think at the circle it goes on as a polygon with an infinite number of sides. Instead of continuing to go straight, the ground acts on the object with a force that increases the normal force and creates the centripetal force needed for the object to go circularly.

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Let's do this using angular momentum as a vector. This should clear up the question on using both axes separately or one new one. The spinning around the y-axis will give an angular momentum in the y-direction: $\vec{L_{y}} = \hat{y} L_{y}$, while the spinning around z-axis gives angular momentum in the z-direction: $\vec{L_{z}} = \hat{z} L_{z}$. We can get ...

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That looks lie a homework style question. So here the necessary hints. The tensors for the base shapes are listed here. It should be clear how to get the tensors if you put two objects together. The same applies if subtracting the inner cone or cylinder for a hollow object. For additional details check the parallel axis theorem

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