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Apply part (a) first to the position

r

(\dot{r})_{S} = (\dot{r})_{S^{'}} + ω \times r .

Applying it again to both terms on the right gives

(\ddot{r})_{S} = (\ddot{r})_{S^{'}} + ω \times (\dot{r})_{S^{'}} + (\dot{ω})_{S^{'}} \times r + ω \times (\dot{r})_{S},

(2)

where

(\dot{ω})_{S} = (\dot{ω})_{S^{'}}

because

ω \times ω = 0

. Substituting the first formula into the last term produces

(\ddot{r})_{S} = (\ddot{r})_{S^{'}} + 2 ω \times (\dot{r})_{S^{'}} + (\dot{ω})_{S^{'}} \times r + ω \times (ω \times r) .

(3)

In the Earth-fixed rotating reference frame, the sideways apparent force is the Coriolis acceleration multiplied by the ski-doo's mass:

F_{side} = - 2 m ω \times v .

(4)

At the South Pole,

ω

is vertical and antiparallel to the local outward normal, while

v

is horizontal, so

∣ F_{side} ∣ = 2 mω v .

(5)

The force points to the left of the direction of travel, which is westward when the ski-doo is moving north after crossing the pole.

Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 16D / c / Solution Created 2026-09-24 Updated 2026-09-25

With

B = 0

and

E = (0, 0, E)

, the spatial Lorentz force equation gives

p_{x} = p_{0}, p_{y} = 0, p_{z} = qEt .

The relativistic energy-momentum relation therefore yields

E (t) = E_{0}^{2} + c^{2} q^{2} E^{2} t^{2}, E_{0} = m^{2} c^{4} + c^{2} p_{0}^{2} .

(2)

Since

v = c^{2} p / E

\overset{z}{˙} = \frac{c ^{2} qEt}{E ( t )} .

(3)

Integration from the initial position

z (0) = 0

gives

z (t) = \frac{E _{0}^{2} + c ^{2} q ^{2} E ^{2} t ^{2} - E _{0}}{qE} .

(4)

Relativistic motion in a constant electric field with transverse momentum Created 2026-09-24 Updated 2026-09-24

For

E = (0, 0, E)

B = 0

, and initial momentum

(p_{0}, 0, 0)

p_{x} = p_{0}, p_{z} = qEt, E (t) = E_{0}^{2} + c^{2} q^{2} E^{2} t^{2} .

With the initial position at the origin,

z (t) = \frac{E ( t ) - E _{0}}{qE}, x (t) = \frac{c p _{0}}{qE} arsinh \frac{c qEt}{E _{0}} .

(2)

Rotating-frame derivative formula Created 2026-09-29 Updated 2026-10-05

If a frame rotates with angular velocity

Ω

relative to an inertial frame, the derivatives of any vector

a (t)

satisfy

(\frac{d a}{d t})_{inertial} = (\frac{d a}{d t})_{rotating} + Ω \times a .

Applying this formula twice to a position gives the Coriolis acceleration, Euler acceleration, and centrifugal acceleration terms.

Rotational deflection of a falling body 2026-10-06

In an Earth-fixed rotating reference frame, the horizontal displacement of a freely falling body receives contributions from Coriolis acceleration and centrifugal acceleration. A short-fall approximation substitutes the unperturbed downward velocity into the Coriolis term and the Earth's surface position into the centrifugal term. If gravitational acceleration excludes centrifugal acceleration, both contributions must be included explicitly.

Velocity Created 2026-09-24 Updated 2026-09-24

Velocity is the time derivative of position.