Съдържание

Условие Отговори Решение

USAPhO 2026, Първи кръг — Задача 22

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Всички материали от USAPhO 2026

2026 $F = ma$ Exam · 1 т.

Указания

25 QUESTIONS - 75 MINUTES

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Use $g = 10\ \mathrm{N/kg}$ throughout, unless otherwise specified.
You may write in this question booklet and the scratch paper provided by the proctor.
This test has 25 multiple choice questions. Select the best response to each question, and use a No.2 pencil to completely fill the box corresponding to your choice. If you change an answer, completely erase the previous mark. Only use the boxes numbered 1 through 25 on the answer sheet.
All questions are equally weighted, but are not necessarily equally difficult.
You will receive one point for each correct answer, and zero points for each incorrect or blank answer. There is no additional penalty for incorrect answers.
You may use a hand-held calculator. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any external references, such as books or formula sheets.
The question booklet, answer sheet and scratch paper will be collected at the end of this exam.
To maintain exam security, do not communicate any information about the questions or their solutions until after February 13, 2026.

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

We acknowledge The US Physics Team coaches and other contributors to this year’s exams:

Tengiz Bibilashvili, Eddie Chen, Orhun Ciftcioglu, Kellan Colburn, Evan Erickson, Harry Kroft, Natalie LeBaron, Mai Qi, Elena Yudovina

Условие

A tricycle of mass $m = 100\ \mathrm{kg}$ is traveling north on a horizontal surface. The geometry of the tricycle is defined as follows:

Wheelbase (distance from front axle to rear axle): $L = 1.0\ \mathrm{m}$
Rear track width (distance between rear wheels): $d = 1.0\ \mathrm{m}$
Center of mass (CM) location: on the longitudinal symmetry axis, a distance $b = 0.4\ \mathrm{m}$ forward of the rear axle
Radius of gyration about the CM: $k = 0.5\ \mathrm{m}$

(The radius of gyration $k$ is defined by $I_{\mathrm{CM}} = mk^2$ , where $I_{\mathrm{CM}}$ is the moment of inertia of the tricycle about a vertical axis through its center of mass.)

Seeing a patch of ice on their right side ahead, the rider panics and slams on the brakes. The front wheel and the left rear wheel slide on dry pavement with a coefficient of kinetic friction $\mu_k = 0.5$ , while the right rear wheel slides on frictionless ice ( $\mu = 0$ ).

This asymmetry in friction forces produces a net torque about the center of mass, causing the tricycle to begin rotating. Calculate the initial angular acceleration $\alpha$ of the tricycle.

(A) $3.0\ \mathrm{rad/s^2}$ , Counter-Clockwise

(B) $3.0\ \mathrm{rad/s^2}$ , Clockwise

(D) $6.0\ \mathrm{rad/s^2}$ , Counter-Clockwise

(E) $3.3\ \mathrm{rad/s^2}$ Clockwise

Отговори

Решение

Consider the forces acting on the three wheels while breaking. The two wheels in contact with the rough surface generates a kinetic friction force opposing the motion (pointing South). Using a symmetry argument we can reduce this down to just the left wheel, which is displaced a lateral distance $d/2$ to the left of the central axis.

Now we calculate the torque generated by the left rear wheel about the CM. The position vector $\vec{r}$ points to the left (West), and the force vector $\vec{F}$ points backward (South). Using the cross product we get:

\vec{\tau}_{left} = (r\hat{\imath}) \times (-F\hat{\jmath})

The force pulls backward on the left side of the vehicle. This creates a twisting moment that rotates the vehicle counter-clockwise (Left).

\tau_{net} = F_{fric} \cdot \frac{d}{2} \quad \text{(Counter-Clockwise)}

Since the torque is non-zero and counter-clockwise, the tricycle acquires an initial angular acceleration $\alpha$ in the counter-clockwise direction.

\alpha \propto \tau_{net} > 0 \implies \text{CCW Rotation}

To find the exact value, we first calculate the moment of inertia ( $I$ ) using the radius of gyration $k$ :

I = mk^2 = 100(0.5)^2 = 25\ \mathrm{kg} \cdot \mathrm{m}^2

Next, we determine the normal forces on the wheels to find the friction magnitude. Summing moments about the rear axle allows us to find the normal force on the front wheel ( $N_f$ ):

\Sigma\tau_{rear} = 0 \implies N_fL = mgb

Plugging in our values above therefore gives us:

N_f(1.0) = (100)(10)(0.4) \implies N_f = 400\ \mathrm{N}

The remaining weight is supported by the two rear wheels. Since the CM is on the symmetry axis, this load is split equally:

N_{rear\_total} = mg - N_f = 1000 - 400 = 600\ \mathrm{N}

N_{left} = \frac{600}{2} = 300\ \mathrm{N}

Now we calculate the kinetic friction force on the left rear wheel:

F_{friction} = \mu_kN_{left} = 0.5(300) = 150\ \mathrm{N}

As we stated previously, the net torque is generated solely by the left rear wheel acting at a lever arm of $d/2$ :

\tau_{net} = F_{friction} \times \frac{d}{2} = 150 \times 0.5 = 75\ \mathrm{N} \cdot \mathrm{m}

Finally, from Newton’s second law for rotation we get:

75 = 25\alpha \implies \alpha = 3.0\ \mathrm{rad/s^2}

Оригинал в Архива: 2026_FMA_exam.pdf · официални решения: 2026_FMA_solutions.pdf