Problem Statement
A train travels east at $v = 100\,\text{km/h}$ at latitude $\lambda = 60°N$. Find the force on the rails from Earth’s rotation (Coriolis effect). Train mass $m = 1000\,\text{t}$.
Given Information
- $v = 100\,\text{km/h}$
- $m = 1000\,\text{t}$
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Coriolis acceleration: $\vec a_{Cor} = -2\vec\omega\times\vec v$
Step 2 — Apply the relevant physical law or equation: Earth’s angular velocity: $\omega = 7.27\times10^{-5}\,\text{rad/s}$
Step 3 — Solve algebraically for the unknown: At latitude $\lambda$, the vertical component of $\vec\omega$ is $\omega\sin\lambda$.
Step 4 — Substitute numerical values with units: Train moves east, vertical component of $\omega$ points upward:
Step 5 — Compute and check the result: $$a_{Cor} = 2\omega\sin\lambda\cdot v = 2\times7.27\times10^{-5}\times\sin60°\times\frac{100000}{3600}$$
$$= 2\times7.27\times10^{-5}\times0.866\times27.8 = 3.50\times10^{-3}\,\text{m/s}^2$$
Step 6: (Directed south in NH, perpendicular to motion = toward right rail)
Worked Calculation
$$a_{Cor} = 2\omega\sin\lambda\cdot v = 2\times7.27\times10^{-5}\times\sin60°\times\frac{100000}{3600}$$
$$= 2\times7.27\times10^{-5}\times0.866\times27.8 = 3.50\times10^{-3}\,\text{m/s}^2$$
$$F = ma_{Cor} = 10^6\times3.50\times10^{-3} \approx 3500\,\text{N}$$
Coriolis acceleration: $\vec a_{Cor} = -2\vec\omega\times\vec v$
Earth’s angular velocity: $\omega = 7.27\times10^{-5}\,\text{rad/s}$
At latitude $\lambda$, the vertical component of $\vec\omega$ is $\omega\sin\lambda$.
Train moves east, vertical component of $\omega$ points upward:
$$a_{Cor} = 2\omega\sin\lambda\cdot v = 2\times7.27\times10^{-5}\times\sin60°\times\frac{100000}{3600}$$
$$= 2\times7.27\times10^{-5}\times0.866\times27.8 = 3.50\times10^{-3}\,\text{m/s}^2$$
(Directed south in NH, perpendicular to motion = toward right rail)
$$F = ma_{Cor} = 10^6\times3.50\times10^{-3} \approx 3500\,\text{N}$$
Answer
$$\boxed{F = ma_{Cor} = 10^6\times3.50\times10^{-3} \approx 3500\,\text{N}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
Leave a Reply