NBPhO 2026

Part i) (1 point)

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

$$L = \sqrt{x^2 + H^2}, \qquad \sin\alpha = x/L.$$

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 $\pi/2 - \alpha$ with the floor) is $u\sin\alpha$. Equating:

$$u\sin\alpha = v \implies \boxed{\,u = \frac{v}{\sin\alpha}.\,}$$

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

$$L\dot{L} = x\dot{x}.$$

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

$$\dot{x} = -\frac{Lv}{x} = -\frac{v}{\sin\alpha},$$ $$u = |\dot{x}| = \frac{v}{\sin\alpha}.$$

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

Part ii) (1 point)

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

$$\dot{L}^2 = \dot{x}^2 + x\ddot{x},$$ $$\ddot{x} = \frac{v^2 - \dot{x}^2}{x} = \frac{v^2 - u^2}{x}.$$

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

$$|\ddot{x}| = \frac{u^2 - v^2}{x} = \frac{v^2(1-\sin^2\alpha)/\sin^2\alpha}{H\tan\alpha}$$ $$\boxed{\,|\ddot{x}| = \frac{v^2}{H}\cot^3\alpha.\,}$$

• 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

$$u = -\dot{x} = -\frac{H}{\cos^2\alpha}\dot{\alpha} = \frac{H\omega}{\cos^2\alpha}.$$

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

$$\omega = \frac{u\cos^2\alpha}{H} = \frac{v\cos^2\alpha}{H\sin\alpha}.$$

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

$$\dot{u} = -\frac{v\cos\alpha}{\sin^2\alpha}\dot{\alpha} = \frac{v\omega\cos\alpha}{\sin^2\alpha}.$$

Substituting $\omega$:

$$|\ddot{x}| = \dot{u} = \frac{v\cos\alpha}{\sin^2\alpha}\cdot\frac{v\cos^2\alpha}{H\sin\alpha} = \boxed{\,\frac{v^2}{H}\cot^3\alpha.\,}$$

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

Part iii) (3 points)

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:

$$T\sin\alpha = M|\ddot{x}| = \frac{Mv^2\cos^3\alpha}{H\sin^3\alpha},$$ $$T = \frac{Mv^2\cos^3\alpha}{H\sin^4\alpha}.$$

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

$$\frac{Mv^2\cos^4\alpha_0}{H\sin^4\alpha_0} = Mg,$$ $$\tan^4\alpha_0 = \frac{v^2}{gH},$$ $$\boxed{\,\alpha_0 = \arctan\sqrt[4]{\frac{v^2}{gH}}.\,}$$

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