 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019-resit/10/b/solution

The radial and angular Euler-Lagrange equations are

(M + m) \overset{r}{¨} = m r \dot{θ}^{2} - M g, \frac{d}{d t} (m r^{2} \dot{θ}) = 0.

J = m r^{2} \dot{θ},

we obtain

(M + m) \overset{r}{¨} = \frac{J ^{2}}{m r ^{3}} - M g .

A positive constant radius requires zero radial acceleration. Hence the circular motion has

r_{0} = (\frac{J ^{2}}{m M g})^{1/3},

provided

J \neq = 0

and

0 < r_{0} < l

. Its angular velocity is

\dot{θ}_{0} = J / (m r_{0}^{2})

. No positive circular radius exists for

J = 0

with

M, g > 0

For later use, the conserved radial energy is

H = \frac{1}{2} (M + m) \overset{r}{˙}^{2} + U_{J} (r), U_{J} (r) = \frac{J ^{2}}{2 m r ^{2}} + M g r - M g l .

U_{J}^{''} (r) = 3 J^{2} / (m r^{4}) > 0

, so its stationary radius is a stable minimum. The tension remains positive in the ideal allowed region: the hanging Newton's second law gives

T_{s} = M (g + \overset{r}{¨}) = M m (g + r \dot{θ}^{2}) / (M + m) > 0

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