More on Mechanics?

Question

More on Mechanics?

1 Answer

Impact of this question

1810 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T12:44:18

See below.

Explanation:

We will be using the so called Euler Lagrange formulation

$\frac{d}{d t} (\frac{\partial L}{\partial {. q}_{i}}) - \frac{\partial L}{\partial q_{i}} = Q_{i}$ $d/dt((partialL)/(partial dot q_i))-(partial L)/(partial q_i)=Q_i$

where $L = T - V$ $L = T-V$ . In this exercise we have $V = 0$ $V=0$ so $L = T$ $L = T$

Calling $x_{a}$ $x_a$ the center of left cylinder coordinate and $x_{b}$ $x_b$ the rigth one, we have

$x_{b} = x_{a} + R cos θ + L cos α$ $x_b=x_a+R costheta+Lcosalpha$

Here $sin α = \frac{R}{L} sin θ$ $sinalpha=R/Lsintheta$ so substituting for $α$ $alpha$

$x_{b} = x_{a} - R cos θ + \sqrt{L^{2} - R^{2} {sin}^{2} θ}$ $x_b=x_a-R costheta + sqrt[L^2 - R^2 sin^2theta]$

now deriving

${. x}_{b} = {. x}_{a} + R sin (θ) . θ - ⎛ ⎜ ⎜ ⎝ \frac{R^{2} cos (θ) sin (θ)}{\sqrt{L^{2} - R^{2} {sin}^{2} (θ)}} ⎞ ⎟ ⎟ ⎠ . θ$ $dot x_b=dot x_a + Rsin(theta) dot theta-((R^2cos(theta)sin(theta))/sqrt(L^2-R^2sin^2(theta)))dot theta$

but

$T = \frac{1}{2} J (ω_{a}^{2} + ω_{b}^{2}) + \frac{1}{2} m (v_{a}^{2} + v_{b}^{2})$ $T=1/2 J (omega_a^2+omega_b^2)+1/2m(v_a^2+v_b^2)$

Here $J$ $J$ is the inertia momentum regarding the mass center. Also,

$v_{a} = {. x}_{a} = R . θ$ $v_a= dot x_a=R dot theta$
$ω_{a} = . θ$ $omega_a = dot theta$

so, after substitutions and calling $ξ (θ) = 1 - \frac{R cos (θ)}{\sqrt{L^{2} - R^{2} {sin}^{2} (θ)}}$ $xi(theta) = 1-(Rcos(theta))/sqrt(L^2-R^2sin^2(theta))$ we have

$T = \frac{1}{2} (J + m R^{2}) (1 + {(1 + sin (θ) ξ (θ))}^{2}) {. θ}^{2}$ $T=1/2(J+mR^2)(1+(1+sin(theta)xi(theta))^2)dot theta^2$

We choosed $θ$ $theta$ as the generalized coordinate. So we will reduce $F$ $F$ actuating in the coordinate $x$ $x$ to an equivalent force in $θ$ $theta$ . This coordinate acts rolling wise so we need a generalized momentum regarding the contact point in the floor, which is

$Q_{θ} = F R (1 + sin θ)$ $Q_(theta)=FR(1+ sintheta)$

The movement equations are obtained after

$(J+mR^2)((1+sin(theta)xi(theta))(cos(theta)xi(theta)+sin(theta)xi'(theta))dot theta^2+(1+(1+sin(theta)xi(theta))^2)ddot theta)=FR(1+sin(theta))$ now solving for $ddot theta$

$ddottheta=(FR(1+sin(theta))-(J+mR^2)(1+sin(theta)xi(theta))(cos(theta)xi(theta)+sin(theta)xi'(theta))dottheta^2)/((J+mR^2)(1+(1+sin(theta)xi(theta))^2))$

Attached two plots. The first shows $theta$ evolution and the second is for $dottheta$

Value of parameters:

$R=0.5,J=1,m=1,L=2$ The applied force is shown in dased red.

enter image source here

Science

Math

Humanities

... and beyond

More on Mechanics?

1 Answer

Explanation:

Related questions

Impact of this question