Solution

Let

e_{i}^{'} (t)

be a Cartesian basis fixed in the rotating frame. A vector

A = A_{i}^{'} e_{i}^{'}

then satisfies

(\frac{d A}{d t})_{S} = \dot{A}_{i}^{'} e_{i}^{'} + A_{i}^{'} \overset{e}{˙}_{i}^{'} = (\frac{d A}{d t})_{S^{'}} + ω \times A,

\overset{e}{˙}_{i}^{'} = ω \times e_{i}^{'}

. Applying this transport formula twice to

r

gives, for constant

ω

a_{S} = a_{S^{'}} + 2 ω \times v_{S^{'}} + ω \times (ω \times r) .

The second and third terms correspond to Coriolis acceleration and centrifugal acceleration in the rotating description.

For the bead, use cylindrical unit vectors

(e_{r}, e_{ϕ}, e_{z})

with

e_{z}

upward and

r = R (sin θ e_{r} - cos θ e_{z}), \overset{e}{˙}_{r} = ω e_{ϕ} .

Projection of

m \overset{r}{¨} = - m g e_{z} + N

along the wire tangent

cos θ e_{r} + sin θ e_{z}

eliminates the smooth-wire normal force

N

and gives

\ddot{θ} = (ω^{2} cos θ - g / R) sin θ .

Solved by gpt-5.6-sol high.

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