NBPhO 2026, Теория и експеримент — Задача 6. Pull-up bar rope

i) (1 т.)

Place the bar at the origin of a coordinate system where $x$ is the horizontal distance from the bar to the weight along the floor, and $H$ is the height of the bar above the floor. The length of the bar-to-weight rope segment is

Method 1 (inextensibility / velocity projection). Since the rope is inextensible, the velocity components of its two endpoints along the rope direction must match: whatever rate one end recedes from a point on the rope, the other end must approach it at the same rate. The puller’s hand pulls the rope along its own direction at speed $v$, so the rope shortens at rate $v$. The weight moves horizontally toward the bar at speed $u$; its velocity component along the rope (which makes angle $\alpha$ with the vertical, hence angle $\frac{\pi}{2} - \alpha$ with the floor) is $u\sin\alpha$. Equating:

Method 2 (implicit differentiation). Differentiating $L^2 = x^2 + H^2$ with respect to time:

With $\dot L = -v$ (the segment shortens at constant rate) and $\sin\alpha = x/L$:

Grading (Method 1).

• Recognising that the velocity components of the rope endpoints along the rope direction must match (rope inextensibility) 0.4 pts
• Identifying the projection $u\sin\alpha$ for the weight’s velocity along the rope 0.3 pts
• Final answer $u = v/\sin\alpha$ 0.3 pts

Grading (Method 2).

• Setting up $L^2 = x^2 + H^2$ with $\dot L = -v$ 0.3 pts
• Correctly differentiating to obtain $L\dot L = x\dot x$ 0.4 pts
• Final answer $u = v/\sin\alpha$ 0.3 pts

ii) (1 т.)

Method 1 (continuing from $L\dot L = x\dot x$). Differentiate again, with $\dot L = -v$ constant (so $\ddot L = 0$):

Since $u^2 > v^2$, we have $\ddot x < 0$: the weight accelerates toward the bar. With $u = v/\sin\alpha$ and $x = H\tan\alpha$:

• Correctly differentiating an incorrect result from the previous part 0.5 pts.

Method 2 (direct differentiation of $u = v/\sin\alpha$). The angle $\alpha$ changes in time as the weight slides; introduce $\omega = -\dot\alpha$ (positive as $\alpha$ decreases). The weight’s position is $x = H\tan\alpha$, so its speed is

Combined with $u = v/\sin\alpha$:

Differentiating $u = v/\sin\alpha$ with respect to time:

Substituting $\omega$:

Grading (Method 1).

• Differentiating $L\dot L = x\dot x$ to obtain $\ddot x = (v^2 - u^2)/x$ 0.5 pts
• Substituting $u$ and $x$ to obtain final answer $a = (v^2/H)\cot^3\alpha$ 0.5 pts

Grading (Method 2).

• Introducing $\omega = -\dot\alpha$ as the angular rate of the rope direction 0.2 pts
• Relating the weight’s speed to $\omega$: $u = H\omega/\cos^2\alpha$ 0.2 pts
• Deriving $\omega = v\cos^2\alpha/(H\sin\alpha)$ from $u = v/\sin\alpha$ 0.2 pts
• Differentiating $u = v/\sin\alpha$ to obtain $\dot u = v\omega\cos\alpha/\sin^2\alpha$ 0.2 pts
• Substituting $\omega$ and obtaining final answer $a = (v^2/H)\cot^3\alpha$ 0.2 pts

iii) (3 т.)

While the weight is on the floor, the only horizontal force on it is the horizontal component of the rope tension $T\sin\alpha$, directed toward the bar. Newton’s second law in the horizontal direction:

Vertically, $T\cos\alpha + N = Mg$, where $N$ is the normal force from the floor. The weight lifts off when $N = 0$, i.e., $T\cos\alpha_0 = Mg$:

Grading.

• Identifying that the only horizontal force on the weight is $T\sin\alpha$ 0.3 pts
• Writing Newton’s second law horizontally: $T\sin\alpha = M|\ddot x|$ 0.3 pts
• Substituting the expression for $|\ddot x|$ from part (ii) 0.2 pts
• Isolating $T$ to obtain $T = Mv^2\cos^3\alpha/(H\sin^4\alpha)$ 0.2 pts
• Correctly setting up the vertical force balance $T\cos\alpha + N = Mg$ 0.5 pts
• Correct lift-off condition $N = 0$, i.e., $T\cos\alpha_0 = Mg$ 0.5 pts
• Algebraic manipulation to $\tan^4\alpha_0 = v^2/(gH)$ 0.5 pts
• Final answer $\alpha_0 = \arctan\sqrt[4]{v^2/(gH)}$ 0.5 pts