- #1

walking

- 73

- 8

I tried solving this problem as follows but the answer I am getting is wrong.

Here is my diagram:

Wight 1 is the one sliding down the wall, the "bar" particle is the centre of mass of the dumbbell, and weight 2 is the one sliding on the floor.

##F_{r1}## is the force of the bar on weight 1, ##F_{w1}## is the force of the wall, ##F_{w2}## is the normal contact force on weight 2, ##F_{r2}## is the force of the bar on weight 2. The directions of these forces was worked out by considering the direction of the net force acting on each weight. Theta is the angle that the bar makes with the horizontal at time t.

I found the following equations by resolving forces, and for the bar, using the fact that ##\mathbf{F}_{ext}=M\mathbf{a}_{cm}##:

##(F_{r1}-F_{r2})\cos \theta = ma_2##

##(F_{r1}-F_{r2})\sin \theta - -2mg - ma_1##

##F_{r2}\cos \theta = ma_2##

##F_{r1}\sin \theta - mg = ma_1##

I solved these to get ##a_1=-\frac{5g}{3}## and ##a_2=-\frac{l}{h}\frac{g}{3}## where ##l## is the component of the bar in the x direction and h is its component in the y direction, both at time t.

Also, based on the definitions of h and l, I got ##h^2+l^2=L^2## which I differentiated with respect to time (since h and l are variable) to get ##h\frac{dh}{dt}+l\frac{dl}{t}=0##. Now I noted that the derivatives in this equation are the speeds of the weights, ie ##\frac{dh}{dt}=v_1## and ##\frac{dl}{dy}=v_2##, since these are the rates at which the weights slide on the wall and floor respectively. Also, the signs are correct because h is getting smaller which agrees with the fact that ##v_1## is negative, and similarly for the ##v_2## equation.

Now when the speeds are equal, let that speed be ##v>0##. Then ##v_1<0## and so ##v_1=-v##, while ##v_2=v##. Plugging into the derivative equation and simplifying we get ##h=l##. So by the pythagorean equation we get ##h=l=\frac{L}{\sqrt{2}}##.

Now differentiating the pythagorean equation again we get ##(\frac{dh}{dt})^2+h\frac{d^2h}{dt^2}+(\frac{dl}{dt})^2+l\frac{d^2l}{dt^2}=0##, ie ##2v^2+\frac{L}{\sqrt{2}}a_1\frac{L}{\sqrt{2}} a_2=0##, and using the values of ##a_1,a_2## we got at the beginning, this becomes ##2v^2=\frac{L}{\sqrt{2}}2g## or finally ##v=\sqrt{\frac{g}{\sqrt{2}}}\sqrt{L}##, which is the wrong answer. I am not sure where I went wrong.

The actual problem and solution are on page 620 here:

https://archive.org/details/TiplerPhysics5Ed.CompleteSolutions1/page/n617

(The answer given there is ##v=\sqrt{g(1-\frac{1}{\sqrt{2}})}\sqrt{L}## which is very similar to what I got.)