Past exam of the physics 1 paper of the University of Oxford / 2019 resit / 10 / c / ii

Solution

J = m l v_{0} /2

, and the initial total energy is

H_{i} = \frac{1}{2} m v_{0}^{2} - \frac{1}{2} M g l .

When the hanging mass reaches the table,

r = l

. Conservation of angular momentum fixes the tabletop tangential speed there to

J / (m l) = v_{0} /2

. At the limiting case of just reaching, radial velocity is zero. Equating the two energies,

\frac{1}{2} m v_{0}^{2} - \frac{1}{2} M g l = \frac{1}{2} m (\frac{v _{0}}{2})^{2},

yields

v_{0, just}^{2} = \frac{4 M g l}{3 m} .

More generally, the boundary radial kinetic energy is

\frac{1}{2} (M + m) \overset{r}{˙} (l)^{2} = \frac{3}{8} m v_{0}^{2} - \frac{1}{2} M g l,

so reaching requires

v_{0}^{2} \geq 4 M g l / (3 m)

, with equality for just reaching and strict inequality for nonzero outward radial speed at contact. This condition is also sufficient:

U_{J}

is strictly convex, and the allowed interval connecting the initial radius to

l

has no intervening higher effective potential barrier. At the threshold the initial radial acceleration is outward and

r = l

is the outer turning point.

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