Solution to Question 37
1. Kinematics of the Stick:
Let the length of the stick be $l$. Let $\theta$ be the angle the stick makes with the horizontal floor.
The position of the bottom end B is $x_B = l \cos\theta$.
The velocity of B is given as $v$.
$$ v = \frac{dx_B}{dt} = -l \sin\theta \frac{d\theta}{dt} $$
$$ \frac{d\theta}{dt} = -\frac{v}{l \sin\theta} $$
Integrating to find the relationship between time $t$ and angle $\theta$ (Assuming the stick starts vertically against the wall at $t=0$, so $\theta = \pi/2$):
$$ \int_{\pi/2}^{\theta} \sin\phi \, d\phi = -\int_0^t \frac{v}{l} \, dt $$
$$ -\cos\theta – (-\cos(\pi/2)) = -\frac{vt}{l} $$
$$ \cos\theta = \frac{vt}{l} \implies t = \frac{l \cos\theta}{v} $$
2. Motion of the Beetle P:
The beetle climbs from B to A with constant speed $u$ relative to the stick. The distance of the beetle from point B at time $t$ is $s = ut$.
The vertical height of the beetle $h$ is given by:
$$ h = s \sin\theta = ut \sin\theta $$
Substituting $t = \frac{l \cos\theta}{v}$:
$$ h(\theta) = u \left( \frac{l \cos\theta}{v} \right) \sin\theta = \frac{ul}{2v} \sin(2\theta) $$
3. Maximizing Height:
We need to maximize $h(\theta)$. The function $\sin(2\theta)$ reaches its maximum value of 1 when $2\theta = 90^\circ$, i.e., $\theta = 45^\circ$.
However, we must check if the beetle is still on the stick at this angle.
Condition 1: Beetle reaches max height while on the stick ($s \le l$).
This occurs if the optimal angle $\theta=45^\circ$ is reached before the beetle hits the top.
At $\theta = 45^\circ$, distance traveled $s = u \cdot t(45^\circ) = u \frac{l \cos 45^\circ}{v} = \frac{ul}{v\sqrt{2}}$.
Constraint: $s \le l \implies \frac{ul}{v\sqrt{2}} \le l \implies u \le v\sqrt{2}$.
If this holds, $h_{max} = \frac{ul}{2v}$.
Condition 2: Beetle reaches the top end A ($s=l$) before $\theta$ becomes $45^\circ$.
This occurs if $u \ge v\sqrt{2}$. The beetle reaches the top (point A) and stays there (or the problem implies “maximum height attained during the process”). The max height is attained exactly when the beetle reaches A.
At this moment, $s=l$, so $t = l/u$.
The angle at this moment is given by $\cos\theta = \frac{v(l/u)}{l} = \frac{v}{u}$.
The height is $h = l \sin\theta$.
$$ h_{max} = l \sqrt{1 – \cos^2\theta} = l \sqrt{1 – \frac{v^2}{u^2}} $$