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This was a practice test question.

Consider the following figure. (If the frictional coefficients are unclear, between A and B we have $k_1 = 0.3$, B and platform is $k_2 = 0.2$. If you further doubts about the diagram ask me in the comments below and I will clarify.)

enter image description here

The blocks are resting on the rotating platform (bold) attached to the rod which is rotating at constant angular velocity $\omega\ \mathrm{rad/s}$. The masses of blocks and distances from the pulley are as given in the figure. The distance of the pulley from the rod (axis) is negligible. Block B is on verge of slipping. Find $\omega$.

What I did :

Let mass of $A = m_1$. Distance of $A$ from axis = $r_1$. Mass of $B = m_2$. Distance of $A$ from axis = $r_2$.

If block $B$ is on verge of slipping $A$ is also on the verge of slipping. Questionable observation. But leads to correct answer. Anyway, we consider a rotating frame attached to the rod. So we give a centrifugal force to $A$ = $m_1 \omega^2 r_1$ and to $B$ equal to $m_2 \omega^2 r_2$. Tension in string $ = T$. We check two cases. Let $B$ slip inwards. Then $A$ slips outwards. So $A$ slips rightwards relative $B$. $A$ was on verge of slipping. Static friction on $A = k_1 m_1 g$. Static friction on $B = k_2 (m_1 + m_2) g$. Then equating forces and solving the system and putting values we get $ \omega = \sqrt{22}\ \mathrm{rad/s}$. The other case ($B$ goes outwards) leads to contradiction.

Though the result is correct, I have a few questions. Why does the fact that $B$ is on the verge of slipping imply that $A$ is also on the verge of slipping. I mean, suppose we just eliminate the rotation and add an external force on $B$ instead. Let the magnitude of force reach $k_2 (m_1 + m_2) g$. Then why is it necessary that the force of static friction between $A$ and $B$ is $k_1 m_1 g$ ? Am I wrong, and is it just that the answer matched, without correct physics?

A detailed explanation of a solution to this problem would be very helpful as I'm a beginner, and I intend to clarify these essential concepts.

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    $\begingroup$ Since $A$ and $B$ are connected by a rope, one will start moving as soon as the other does. Thus they are both on the verge of slipping at the same time. $\endgroup$ – Brian Moths Aug 6 '14 at 18:48
  • $\begingroup$ Can we prove that symbolically? My attempt was like this : Since the blocks are at rest $ 2T + f = F$ where $F$ is outward on $B$ and $f$ is friction from ground on $B$. Now as $A$ is at rest $T = f'$ where $f'$ is friction on $A$. We get $F = 2f' + f$. Can we show that if $f = k_2 (m_1 + m_2)g $, $f' = k_1m_1 g$. Note I omitted the rotation and instead added an external force on $B$. $\endgroup$ – Mriganka Basu Roy Chowdhury Aug 7 '14 at 7:20
  • $\begingroup$ You would have to add a similar force on $A$. Adding $f$ to "mimic" the centrifugal effect on $B$ is ok, but this is also present on $A$. Also, why $2T$? $\endgroup$ – rmhleo Aug 8 '14 at 6:48
  • $\begingroup$ I meant that $f$ was the friction on $B$. And because I was trying to analyze why $A$ would slip if $B$ slipped, I omitted an external force on $A$ in the reduced case. Besides $2T$ refers to the tensional force on the system, $T$ on A, $T$ on B, for a total of $2T$ on the $A+B$ system. $\endgroup$ – Mriganka Basu Roy Chowdhury Aug 8 '14 at 13:23
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Apparently the only thing you are missing to understand these results is that there is a constant relation between the reaction of the supporting surface, which generally equals the weight of the body $mg$, and the friction opposing its movement $Fr = \mu mg$ where $\mu$ is the friction coefficient. So there is the answer of why to move $B$ you need to exert that force which is related to the reaction of the platform to their combined weight $(m_A + m_B)g$, whereas to move $A$ you just have to overcome the friction on $A$ which is related to the reaction of its weight $m_Ag$.

Now, to solve the problem you don't need to start knowing extra information, you just need to correctly apply Newtons' Laws. Let us write the sum of forces on $A$:

$$F_c(A) - T_A - F_r(AB) = m_A \dot v_A$$

where only components on radial direction are put since is clear that there is no movement in any other direction, therefore all other components compensate each other. Now we write these for $B$:

$$F_c(B) - T_B - F_r(B) - F_r(BA) = m_B \dot v_B$$

Notice that in these equations I chose positive sign for forces along the outwards sense, but this is not important since the final solution's sign will say where the movement goes: outwards if positive, inwards if negative. Also notice that friction's sign is chosen opposite to movement, so since $\dot v$ has positive sign, the corresponding $F_r$ has the negative sign, meaning is opposed to movement direction.

Now the fact that bodies are tided means that: $$\dot v_A = - \dot v_B$$ $$T_A = T_B$$ $$F_r(AB) = - F_r(BA)$$ putting this in the above equations and subtract one to the other we get:

$$F_c(A) - F_c(B) + F_r(B) = (m_A + m_B)\dot v_A$$

where putting the complete form of each force, and clearing $\dot v_A$ gives:

$$\omega^2 \frac{m_A r_A - m_B r_B}{m_A + m_B} - k_2 g = \dot v_A$$ which gives for your values:

$$ 0.3 g - 0.2 \omega^2 = \dot v_A$$

and for static behaviour ($\dot v_A = 0$) you get the frequency for this conditions $\omega_s$ (I get about 4 times less your value though). But also from this equation you see what happens in dynamical conditions: $v_A$ (and therefore $v_B$) depends linearly on $\omega^2$ thus slower rotation speeds ($\omega < \omega_s$) give you a slide of $A$ outwards, while higher rotations ($\omega > \omega_s$) yield an opposite behaviour.

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    $\begingroup$ Won't we consider the the force applied by $A$ on $B$, as $B$ applies a frictional force on $A$ ? Or am I missing anything? $\endgroup$ – Mriganka Basu Roy Chowdhury Aug 7 '14 at 7:21
  • $\begingroup$ Yes, sorry. Thanks for your comment. Now $\omega_s$ is very different. How did you get your value? $\endgroup$ – rmhleo Aug 8 '14 at 6:42
  • $\begingroup$ One more question. Isn't $T_A = T_B$? Both of the forces act "toward the pulley", as the string "tries to hold itself together"? I don't really know, I might be wrong. But if this changes, my answer ($ \sqrt{22} $) would also change. $\endgroup$ – Mriganka Basu Roy Chowdhury Aug 8 '14 at 13:28
  • $\begingroup$ Yes, according to the reference frame chosen they should have the same sign. I still don't see how that makes that value. $\endgroup$ – rmhleo Aug 9 '14 at 8:42

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