Question #b8bfd

1 Answer
Oct 20, 2016

Answer:

#M = 2m(1-r/R)#

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 sintheta=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 costheta-f_2=0#

Vertically

#-m g+ f_c sintheta = 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(r+2r sintheta)+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 #costheta=(R-r)/r# and #sintheta = 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 = 2m(1-r/R)#

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 = << delta p, f >>#

#delta p_1= r delta alpha hat i#

#delta p_2 = delta p_1 + 2r(cos(theta-delta alpha)-cos(theta)) hat i + 2r(sin(theta-delta alpha)-sin(theta))hat j# so

#delta p_2 approx r delta alpha hat i + 2r sintheta delta alpha hat i -2rcostheta delta alpha hat j#

Now calling #p_T = l cos alpha hati + l sin alpha hatj# 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 = << delta p_1, -m g hat j>> + << delta p_2,-m g hat j >> + << delta p_T, -M g hat j>> = 0# or

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

#M = (2 m r cos theta)/(l cos(alpha))# but #l cos(alpha)=R# and #costheta = (R-r)/r# so finally

#M = 2m(1-r/R)#

Note:
The detailed calculation of #deltap_2# follows. As we know

#p_1 = (r-2R)hati+r hat j# and
#p_2=p_1+2r costheta hati+2r sintheta hatj#
After a rigid rotation of #delta alpha# counterclockwise the #p_2# new coordinates are

#p'_2=p'_1+2rcos(theta-deltaalpha)hati+2rsin(theta-deltaalpha)hatj#

then

#deltap_2 = p'_2-p_2#

but

#cos(theta-deltaalpha)=costhetacosdeltaalpha+sinthetasindeltaalpha#
#sin(theta-deltaalpha)=sintheta cosdeltaalpha-costhetasindeltaalpha#

and also

#cosdeltaalpha approx 1# and #sindeltaalpha approx deltaalpha#

Finally

#delta p_2 = r delta alpha hat i + 2r sintheta delta alpha hat i -2rcostheta delta alpha hat j#