WPE CYU 5

Solution to Question 5

Problem Analysis:

We have a system of three identical blocks of mass $m$. The middle block is initially held such that the cord is straight (horizontal). Upon release, the middle block falls, pulling the side blocks inward and upward. We need to find the acceleration of the middle block at the instant the angle between the two segments of the cord becomes $90^\circ$.

Step 1: Geometry and Velocity Relations

Let $l$ be the half-distance between the pulleys. Let $y$ be the vertical displacement of the middle block. The length of the hypotenuse string segment is $x = \sqrt{y^2 + l^2}$.

The total length of the string on one side (connecting middle mass to one side mass) is constant. If the middle mass moves down by distance $y$, the hypotenuse increases from $l$ to $\sqrt{y^2+l^2}$. The side mass rises by an amount $h$ equal to the increase in hypotenuse length:

$$ h = \sqrt{y^2 + l^2} – l $$

Condition: The segments make a right angle. The semi-vertical angle is $\theta = 45^\circ$. From geometry, $\tan \theta = l/y \implies \tan 45^\circ = 1$, so $y = l$.

Constraint Relationship (Velocity):

Differentiating the string length relation ($x + z = C$, where $z$ is the depth of side mass):

$$ v_{side} = -\frac{dx}{dt} = -\frac{d}{dt}\sqrt{y^2+l^2} = -\frac{y}{\sqrt{y^2+l^2}} v_{mid} = -v_{mid} \cos\theta $$

Considering magnitudes: $v_{side} = v \cos\theta$. At $\theta = 45^\circ$, $v_{side} = v/\sqrt{2}$.

Step 2: Conservation of Energy

We apply the Work-Energy theorem from the initial state (horizontal, $y=0$) to the final state ($y=l$).

• Decrease in PE of middle block: $mgy = mgl$.
• Increase in PE of two side blocks: $2mg h = 2mg(\sqrt{l^2+l^2} – l) = 2mgl(\sqrt{2}-1)$.
• Kinetic Energy: $KE = \frac{1}{2}mv^2 + 2(\frac{1}{2}m v_{side}^2)$.

Energy Equation:

$$ mgl = 2mgl(\sqrt{2}-1) + \frac{1}{2}mv^2 + m\left(\frac{v}{\sqrt{2}}\right)^2 $$ $$ gl – 2gl(\sqrt{2}-1) = \frac{1}{2}v^2 + \frac{1}{2}v^2 = v^2 $$ $$ v^2 = gl [ 1 – 2\sqrt{2} + 2 ] = gl(3 – 2\sqrt{2}) $$

Step 3: Acceleration Constraint

Differentiating the velocity relation $v_{side} = v \cos\theta$ with respect to time:

$$ a_{side} = \frac{d}{dt}(v \cos\theta) = a \cos\theta + v (-\sin\theta) \dot{\theta} $$

We know $\dot{\theta} = -\frac{v \sin\theta}{x}$. Substituting this:

$$ a_{side} = a \cos\theta + \frac{v^2 \sin^2\theta}{x} $$

At $\theta = 45^\circ$, $x = l\sqrt{2}$, $\cos\theta = \sin\theta = 1/\sqrt{2}$:

$$ a_{side} = \frac{a}{\sqrt{2}} + \frac{v^2 (1/2)}{l\sqrt{2}} = \frac{a}{\sqrt{2}} + \frac{v^2}{2\sqrt{2}l} $$

Step 4: Dynamics (Newton’s Laws)

Let $T$ be the tension.
Side Block (moving up):

$$ T – mg = m a_{side} \implies T = m(g + a_{side}) $$

Middle Block (moving down):

$$ mg – 2T \cos\theta = ma $$

Substitute $T$ and $\cos 45^\circ = 1/\sqrt{2}$:

$$ mg – 2[m(g + a_{side})] \frac{1}{\sqrt{2}} = ma $$ $$ g – \sqrt{2}(g + a_{side}) = a $$ $$ g(1-\sqrt{2}) = a + \sqrt{2} a_{side} $$

Substitute the constraint expression for $a_{side}$:

$$ g(1-\sqrt{2}) = a + \sqrt{2} \left( \frac{a}{\sqrt{2}} + \frac{v^2}{2\sqrt{2}l} \right) $$ $$ g(1-\sqrt{2}) = a + a + \frac{v^2}{2l} = 2a + \frac{v^2}{2l} $$

Now substitute $v^2 = gl(3 – 2\sqrt{2})$:

$$ g(1-\sqrt{2}) = 2a + \frac{gl(3 – 2\sqrt{2})}{2l} $$ $$ g – g\sqrt{2} = 2a + \frac{3g}{2} – g\sqrt{2} $$ $$ g – \frac{3g}{2} = 2a $$ $$ -\frac{g}{2} = 2a \implies a = -\frac{g}{4} $$