EnergyHard

Spring, rod and body (roto-translation)

A homogeneous rod and a point body, each of mass m₁ = m₂ = 2 kg, lie on a smooth horizontal plane. A spring (k = 40000 N/m) compressed by Δx = 5 cm sits between one end of the rod and the body. Find the speed v of the body after release.

Given data

m₁ = m₂ = 2 kgk = 40000 N/mΔx = 0.05 m

Review the theory: Energia e Lavoro

Steps

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The compressed spring stores elastic potential energy. On release this is the only available energy and it will all convert into kinetic energy of the system: that's why I compute it first. The energy of a spring compressed by Δx is E = ½·k·Δx².
Write your working — the numerical result must appear in the text J
The plane is smooth, so no external forces/torques act: energy, linear momentum and angular momentum are all conserved. The rod does not only translate but also rotates (roto-translation), so the elastic energy is shared between body and rod. Imposing the three conservation laws gives ½kΔx² = (5/2)·m·v², hence v = √(kΔx²/(5m)).

Full worked solution

The compressed spring stores elastic potential energy. On release this is the only available energy and it will all convert into kinetic energy of the system: that's why I compute it first. The energy of a spring compressed by Δx is E = ½·k·Δx².
$E = \tfrac12 k\,\Delta x^2 = \tfrac12\cdot 40000\cdot 0.05^2$
$E = \frac12\cdot40000\cdot0.0025 = \mathbf{50\,J}$ .
The plane is smooth, so no external forces/torques act: energy, linear momentum and angular momentum are all conserved. The rod does not only translate but also rotates (roto-translation), so the elastic energy is shared between body and rod. Imposing the three conservation laws gives ½kΔx² = (5/2)·m·v², hence v = √(kΔx²/(5m)).
$v = \sqrt{\frac{k\,\Delta x^2}{5m}} = \sqrt{\frac{40000\cdot 0.0025}{5\cdot 2}}$
$v = \sqrt{100/10} = \sqrt{10} \approx \mathbf{3.16\,m/s}$ . The 5/2 factor comes from $I = m\ell^2/12$ .

Result:

v \approx 3.16

m/s

Spring, rod and body (roto-translation)

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