 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1/section-c/c12/a/solution

Power is transmitted at rate

P = C Ω

, so

C = \frac{4.0 \times 1 0 ^{4}}{300} N m = 133.3 N m .

For

a = 0.040 m

L = 3.0 m

and

G = 80 \times 1 0^{9} Pa

J_{p} = 4.021 \times 1 0^{- 6} m^{4}

. Hence

ϕ = \frac{C L}{G J _{p}} ≃ 1.24 \times 1 0^{- 3} rad = 0.071 2^{\circ} .

The density does not enter this static torsional deformation.

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