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Do I need to use the angular velocity vector in the rotating or inertial reference frame for this? Yes. You can do it either way. I start with the expression that relates the time derivative of a vector quantity $\boldsymbol u$ in the inertial and rotating frames: $$\left(\frac {d\boldsymbol u}{dt}\right)_I = \left(\frac {d\boldsymbol u}{dt}\right)_R ... 4 To add to David Hammen's answer on the question: When numerically integrating this, together with Euler's equation of rotation, is there a way to ensure that the determinant of R remains equal to one (otherwise \vec{x}(t) will also be scaled)? Method 1 Dumb But Effective Naïve Multiplication Whilst you are getting up to speed with more ... 3 The mistake is in the second line, in the calculation of the differential mass element. The differential mass element in this case is a disc, of radius r where r = R \cos\theta as you have correctly used. However, the thickness of this differential disc is NOT  R d\theta but Rd\theta cos\theta. Try to wrap your head around this. Rd\theta is the ... 2 Any rigid body in motion can be described as rotating about in instantaneous axis of rotation (IAR) and translating along the same axis at the same time. Example/Proof A rigid body in moving and at time instant a point A riding on the rigid body has position vector \vec{r}_A and instantaneous linear velocity \vec{v}_A at A. The whole body is rotating ... 2 Notice that you have implicitly chosen to measure angular momentum about the axle of the platform. That means that all the forces exerted by the axle on the platform are applied through the axis for rotation, meaning the torque they exert is$$\text{force} \times \text{lever arm} = F \times 0 = 0\,.$$And there are no other forces present expect those ... 1 I can't really follow your work, but here's one way to do it. Take the axis to be the z axis. The distance of a point \left(r,\theta,\phi\right) in the sphere from the z axis is r \sin \theta, so$$ \begin{eqnarray} I &=& \int dV \rho \left(r \sin \theta\right)^2 \\ &=& \rho \int_0^{2\pi} d\phi \int_0^\pi d\theta \int_0^R dr \ r^2 ...

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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 ...

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The velocity jacobian is $\vec{\omega}_B = J\, \dot{q}$ with $q=(\phi,\theta,\psi)$. This is used to transform between the generalized forces/torques $Q$ and the vector torques $\vec{M}_B=(\tau_x^B,\tau_y^B,\tau_z^B)$ $$Q = J^\intercal \vec{M}_B$$ The power through the joint is $$Q \cdot \dot{q} = Q^\intercal \dot{q} = \left(J^\intercal ... 1 1- Is it right to say "the motion of the robot can be described as a transitional motion of center of mass plus a rotational motion about that point?" Pick a point on (or off) your robot; pick any point. The motion can always be described in terms of the translational motion of that point plus a rotational motion about that point In general, the ... 1 It seems to me what you're asking is pretty simple. You say you can control the angular velocity of each wheel. That, times the wheel radius, give you the forward velocity of each wheel on the ground. That tells you the robot's forward speed (the average of the forward speeds of the wheels), and it tells you the rate at which the robot is turning (the ... 1 As one of the comments mentions, it is simpler to consider a linear case. Dropping a body of mass m on one moving with mass M and velocity v is essentially considered the instantaneous transformation M \to M + m. Momentum must be conserved in the collision, but the mass of the system effectively increases, producing a smaller kinetic energy:$$ ...

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Energy is conserved, but if you ignore some kinds of energy then it will look like it isn't conserved. You can imagine a really big disk with some radial pointing two by fours attached at the one o'clock, two o'clock etcetera positions then attach springs to each two by four with the spring pointing in the clockwise/counter-clockwise directions. Add a nice ...

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Does a body always rotate purely about its center of mass? Well, that depends. The first assumption you need is that the body is rigid. Violate this assumption and all bets are off the table because you can't even necessarily classify all motions as "rotations": for example if a long thin board starts twisting sinusoidally into/out-of a helix shape, ...

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Say the projectile was thrown with a velocity $v$ at an angle $\theta$ with respect to the horizontal. We ignore all friction effects (air drag, side winds). Define a coordinate system with a vertical $y$-axis, a horizontal $x$-axis and the point of origin $O$ the point from which the projectile starts its trajectory. The trajectory can now be decomposed in ...

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