Analysis of the motion of a rod sliding against a wall using the Alternate Method (Energy & Force Balance).
Step 1: Conservation of Energy
As noted in the problem statement, “Only the upper ball loses P.E.” relative to the initial position (where the rod is vertical). The lower ball moves horizontally along the floor, so its gravitational potential energy does not change.
Let the rod rotate with angular velocity $\omega$. The kinetic energy of the system can be written as rotational kinetic energy regarding the bottom point as the instantaneous pivot:
Simplifying for $\omega^2$:
$$ \omega^2 = \frac{2g}{l}(1 – \sin \theta) \quad \text{…(i)} $$
Step 2: Force Analysis (Condition for Separation)
We consider the forces acting on the upper mass along the line of the rod. At the moment the rod loses contact with the wall, the normal force becomes zero. The component of gravity along the rod provides the necessary centripetal force.
Component of gravity along the rod: $mg \sin \theta$
Step 3: Solving for the Angle
Now, we substitute the value of $\omega^2$ from equation (i) into equation (ii):
$$ mg \sin \theta = m \left[ \frac{2g}{l}(1 – \sin \theta) \right] l $$
Canceling $m$, $g$, and $l$:
$$ \sin \theta = 2(1 – \sin \theta) $$
$$ \sin \theta = 2 – 2 \sin \theta $$
$$ 3 \sin \theta = 2 $$
Step 4: Finding Velocity (u)
Using the derived angle, we first find the angular velocity $\omega$:
$$ \omega = \sqrt{\frac{2g}{3l}} $$
The horizontal velocity $u$ of the bottom mass is related to angular velocity by $u = \omega r_{\perp}$, where the perpendicular distance is $y = l \sin \theta$. (Or simply derived from $x = l \cos \theta \implies \dot{x} = -l \sin \theta \cdot \omega$).
$$ u = \omega l \sin \theta $$
$$ u = \sqrt{\frac{2g}{3l}} \cdot l \cdot \frac{2}{3} $$
Solution
Let the rod have length $l$. Let $\theta$ be the angle the rod makes with the vertical.
Initially, the rod is vertical ($\theta = 0$). As it slides, mass A moves down and mass B moves right.
1. Velocity Relations:
Let $v$ be the velocity of the lower ball B. $v = v_x$.
Using the constraint $x^2 + y^2 = l^2$:
$$ 2x\dot{x} + 2y\dot{y} = 0 \implies v_y = – \frac{x}{y} v_x = – v \tan \theta $$
Total Kinetic Energy ($K$):
$$ K = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 = \frac{1}{2}mv^2 (1 + \tan^2 \theta) = \frac{1}{2}mv^2 \sec^2 \theta $$
2. Conservation of Energy:
Only the upper ball loses potential energy. The lower ball moves horizontally on the floor.
Loss in PE = $mg(l – y) = mg(l – l \cos \theta) = mgl(1 – \cos \theta)$.
Equating Loss in PE to Gain in KE:
$$ mgl(1 – \cos \theta) = \frac{1}{2}mv^2 \sec^2 \theta $$
Solving for $v^2$:
$$ v^2 = 2gl (1 – \cos \theta) \cos^2 \theta $$
3. Condition for Leaving the Wall:
The upper ball leaves the wall when the normal force from the wall becomes zero. This corresponds to the moment when the horizontal acceleration of the Center of Mass changes behavior, or more simply, when the velocity of the lower ball B is maximized .
To maximize $v$, we maximize $v^2$. Let $u = \cos \theta$. Function to maximize:
$$ f(u) = (1 – u)u^2 = u^2 – u^3 $$
Differentiating w.r.t $u$:
$$ \frac{df}{du} = 2u – 3u^2 = 0 $$
$$ u(2 – 3u) = 0 \implies u = \frac{2}{3} $$
So, the ball leaves the wall when $\cos \theta = 2/3$.
4. Calculate Maximum Speed:
Substitute $\cos \theta = 2/3$ back into the velocity equation:
$$ v^2 = 2gl \left(1 – \frac{2}{3}\right) \left(\frac{2}{3}\right)^2 $$
$$ v^2 = 2gl \left(\frac{1}{3}\right) \left(\frac{4}{9}\right) $$
$$ v^2 = \frac{8gl}{27} $$
Answer: $v = \sqrt{\frac{8gl}{27}}$