 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-1/section-c/c8/a/solution

The original apparatus has two small masses at distance

L /2

from the suspension axis. Neglecting the rod's moment of inertia, their total moment of inertia is

I = 2 m (L /2)^{2} = m L^{2} /2

, consistent with the supplied torsion pendulum period. Treat

m

and

L

as exact because no uncertainties are specified. Rearranging the period relation gives

κ = \frac{2 π ^{2} m L ^{2}}{T ^{2}} = \frac{2 π ^{2} ( 0.038 kg ) ( 0.100 m ) ^{2}}{( 498.2 s ) ^{2}} = 3.02208 \times 1 0^{- 8} N m ra d^{- 1} .

The torsion constant is a restoring torque per unit angle. In dimensional calculations the radian is dimensionless.

σ_{κ} = \frac{d κ}{d T} σ_{T} = \frac{2 κ}{T} σ_{T}, \frac{σ _{κ}}{κ} = \frac{2 ( 6.0 )}{498.2} = 0.02409.

Consequently

κ = (3.022 \pm 0.073) \times 1 0^{- 8} N m ra d^{- 1} .

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