RBD BYU 28

Step 1: Initial Tension ($T_i$)

Before the string is cut, the system is in equilibrium. We calculate the tension in the left string ($T_i$) by taking torque about the right end (point B) to eliminate $T’$.

Total length is $3l$. Mass $m$ is at distance $2l$ from B. Mass $2m$ is at distance $l$ from B.

$$ \Sigma \tau_B = 0 $$ $$ T_i(3l) = mg(2l) + 2mg(l) $$ $$ 3T_i = 4mg \implies T_i = \frac{4mg}{3} $$

Step 2: Dynamics After Release

When the right string is cut, the tube rotates about point A. Let the angular acceleration be $\alpha$. The linear acceleration is $a = r\alpha$.

• Acceleration of mass $m$ (at $l$): $a_1 = l\alpha = a_0$
• Acceleration of mass $2m$ (at $2l$): $a_2 = 2l\alpha = 2a_0$

Since the tangential acceleration $a > g$ for the outer mass (as we will see), the rod must push it down. Conversely, the rod must hold the inner mass up. Let $T$ be the magnitude of the force parameter from the massless rod equations.

Equations of motion for the beads:

1. For mass $m$: Rod pulls up with force $2T$ (effective). $$ mg – 2T = m(a_0) \quad \dots(1) $$ 2. For mass $2m$: Rod pushes down with force $T$. $$ 2mg + T = 2m(2a_0) = 4ma_0 \quad \dots(2) $$

Step 3: Solving for Final Tension ($T_f$)

Solving the system of equations (1) and (2). From (1), $2T = mg – ma_0 \Rightarrow T = \frac{mg – ma_0}{2}$. Substitute into (2):

$$ 2mg + \frac{mg – ma_0}{2} = 4ma_0 $$ $$ 4mg + mg – ma_0 = 8ma_0 $$ $$ 5mg = 9ma_0 \implies a_0 = \frac{5g}{9} $$

Now find the tension parameter $T$:

$$ T = \frac{mg – m(5g/9)}{2} = \frac{4mg/9}{2} = \frac{2mg}{9} $$

Force Balance on the Rod: Since the rod is massless, the upward tension from the string $T_f$ must balance the net forces from the beads. The rod experiences $2T$ downward (reaction from inner bead) and $T$ upward (reaction from outer bead).

$$ T_f + T – 2T = 0 \implies T_f = T = \frac{2mg}{9} $$