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## New answers tagged dynamics

0

The relationship between points B and C $$\vec{v}_C = \vec{v}_B + \vec{W}_{BC} \times \vec{r}_{BC} \tag{eq.1}$$ Additionally you know that $$\vec{v}_D = \vec{v}_C + \vec{W}_{CD} \times \vec{r}_{CD} \tag{eq.2}$$ However you already know that: $v_B = \begin{bmatrix}0 \\-3j\\0 \end{bmatrix}$ $v_D = \begin{bmatrix}0 \\0\\0 \end{bmatrix}$ If you do the calcs ...

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Find the transfer function $\delta_r(t)\to \psi (t)$ and $\delta _r(t)\to \beta (t)$. Divide the first transfer function by the second to get the transfer function $\beta (t)\to \psi (t)$. These calculations can be tedious. I used Mathematica to get the following result. ssm = StateSpaceModel[{p'[t] == -11.450 p[t] + 2.7185 r[t] - 19.4399 \[Beta][t] + ...

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Why dont you construct a state-space model from the given equations (which happen to be in the appropriate form) and either continue using the state space model or convert it to a transfer function in matlab. A continuous-time state-space model looks like this: $$\dot{x} = Ax + Bu$$ $$y = Cx+Du$$ Where $x$ and $u$ are vectors and $A$ and $B$ matrices. $$\... 0 Knowing that the derivative of a vector in frame A can be written as: And similarly, the derivative of the same vector written in the B frame can be written as: Substituting the second equation in the first one we have: For the above to be true, either the vector being differentiated is a null vector (which is a trivial solution) or: 0 H-infinity control uses just as any other controller the outputs to compute a new input. Internally, it does store some kind of linear combination of previous outputs and inputs that assure the resulting control signal stabilizes the system and achieves the control goal. This can be compared to how a LQR + Observer works, the observer estimates something ... 0 This is in the right track but not entirely.$$T_1+V_1=T_2+V_2$$I found two problems one with the initial position. Let's assume position is h=0 initially, and the mass travels by S=1m. So the initial energy can be thought of as zero. the other is with an added g in the rotational energy of the pulley. It should be \frac{1}{2} Iw^2  instead of \frac{1}{... 1 Again your basic formula is just fine. \tau=I_o \cdot \alpha,. Therefore$$ \alpha = \frac{\tau}{I_o}$$The reason why you are getting such a high acceleration is due to the very high torque \tau with respect to the reported mass moment of inertia and radius. I would crosscheck the values regarding the mass moment of inertia because given the radius and ... 1 The CG is found by assuming the mass of the cone as 1/3m and the cylinder m, then$$ \bar{X}= \frac{2r*m+5r*m/3}{4/3m} =r*11/4 $$The cone side length (like the sharp tip of a pencil) is 5r, the side of a 3,4,5 triangle, and the interior half tip angle is 36.87 degrees. The CG is 5.25r from the tip of the cone and at rotation to rest on the side of the ... 3 Given the geometry of the object, the center of gravity is very close to the surface between the cylinder and the cone. More specifically the distance between the center of gravity and the base of the cone is \frac{1}{4}r. Basically what you need to prove is that the angle of the cone  \phi is such, so that when the object is tilted the weight crosses ... 0 Why would you think it should be more complicated. Angular momentum is a vector, comparable to linear momentum. And all the laws of preservation of momentum apply to it. In the most basic form if you have a particle with a mass of m moving in an arc about an axis is r = distance of the particle from the axis m= mass v =speed of the particle P = linear ... 0 yes, a disk rotating about its center axis with moment of inertia I and angular velocity \omega has angular momentum L:$$L = I \cdot \omega it is as simple as that.

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