# Question b8bfd

Oct 20, 2016

$M = 2 m \left(1 - \frac{r}{R}\right)$

#### Explanation:

Considering ${B}_{1}$ and ${B}_{2}$ with ${B}_{1}$ at the botton, its static equilibrium is attained with the forces

For ${B}_{1}$

Horizontally

${f}_{1} - {f}_{c} \cos \theta = 0$

Vertically

$- m g + {n}_{1} - {f}_{c} \sin \theta = 0$

here ${f}_{1}$ is the horizontal contact force with the tube wall and ${f}_{c}$ is the contact force between ${B}_{1}$ and ${B}_{2}$ and ${n}_{1}$ the floor contact force.

For ${B}_{2}$

Horizontally

${f}_{c} \cos \theta - {f}_{2} = 0$

Vertically

$- m g + {f}_{c} \sin \theta = 0$

here ${f}_{2}$ is the horizontal contact force with the tube ball.

Now the global equilibrium states that the momentum of the forces acting over the tube must sum zero or

${f}_{1} r - {f}_{2} \left(r + 2 r \sin \theta\right) + M g R = 0$

Here we suppose that the tube has the right bottom extremum as possible rotational pivot, at the knock down limit, so the support normal forces are concentrated at this point.

Here $\cos \theta = \frac{R - r}{r}$ and $\sin \theta = \frac{\sqrt{2 r R - {R}^{2}}}{r}$

Solving the system of equations

{ (f_1-f_c cos theta=0), (-m g + n_1 -f_c sintheta=0), (f_c costheta-f_2=0), (-m g +f_c sintheta = 0), (f_1 r-f_2(r+2r sintheta)+M g R =0) :}

for ${f}_{1} , {f}_{2} , {f}_{c} , {n}_{1} , M$ we obtain

f_1 = (costheta g m)/sintheta, f_2 = (costheta g m)/sintheta, f_c = (g m)/sintheta, n_1 = 2 g m, M = (2 costheta m r)/R#

so

$M = 2 m \left(1 - \frac{r}{R}\right)$

d'Alembert's principle analysis will come later.

Analysing the virtual work of each body ${B}_{1} , {B}_{2}$ and $T$ (for tube) we have $\delta w = \left\langle\delta p , f\right\rangle$

$\delta {p}_{1} = r \delta \alpha \hat{i}$

$\delta {p}_{2} = \delta {p}_{1} + 2 r \left(\cos \left(\theta - \delta \alpha\right) - \cos \left(\theta\right)\right) \hat{i} + 2 r \left(\sin \left(\theta - \delta \alpha\right) - \sin \left(\theta\right)\right) \hat{j}$ so

$\delta {p}_{2} \approx r \delta \alpha \hat{i} + 2 r \sin \theta \delta \alpha \hat{i} - 2 r \cos \theta \delta \alpha \hat{j}$

Now calling ${p}_{T} = l \cos \alpha \hat{i} + l \sin \alpha \hat{j}$ where $l$ is the distance between the pivot point and the tube center of mass, we have

$\delta {p}_{T} = - l \sin \alpha \delta \alpha \hat{i} + l \cos \alpha \delta \alpha \hat{j}$

Finally

$\delta w = \left\langle\delta {p}_{1} , - m g \hat{j}\right\rangle + \left\langle\delta {p}_{2} , - m g \hat{j}\right\rangle + \left\langle\delta {p}_{T} , - M g \hat{j}\right\rangle = 0$ or

$2 g m r \cos \theta - g l M \cos \alpha = 0$ solving for $M$

$M = \frac{2 m r \cos \theta}{l \cos \left(\alpha\right)}$ but $l \cos \left(\alpha\right) = R$ and $\cos \theta = \frac{R - r}{r}$ so finally

$M = 2 m \left(1 - \frac{r}{R}\right)$

Note:
The detailed calculation of $\delta {p}_{2}$ follows. As we know

${p}_{1} = \left(r - 2 R\right) \hat{i} + r \hat{j}$ and
${p}_{2} = {p}_{1} + 2 r \cos \theta \hat{i} + 2 r \sin \theta \hat{j}$
After a rigid rotation of $\delta \alpha$ counterclockwise the ${p}_{2}$ new coordinates are

$p {'}_{2} = p {'}_{1} + 2 r \cos \left(\theta - \delta \alpha\right) \hat{i} + 2 r \sin \left(\theta - \delta \alpha\right) \hat{j}$

then

$\delta {p}_{2} = p {'}_{2} - {p}_{2}$

but

$\cos \left(\theta - \delta \alpha\right) = \cos \theta \cos \delta \alpha + \sin \theta \sin \delta \alpha$
$\sin \left(\theta - \delta \alpha\right) = \sin \theta \cos \delta \alpha - \cos \theta \sin \delta \alpha$

and also

$\cos \delta \alpha \approx 1$ and $\sin \delta \alpha \approx \delta \alpha$

Finally

$\delta {p}_{2} = r \delta \alpha \hat{i} + 2 r \sin \theta \delta \alpha \hat{i} - 2 r \cos \theta \delta \alpha \hat{j}$