Caterpillar tank

The exercise of the tank to caterpillar was an old traditional exercise of the baccalaureat, in mechanics of the simple systems:

that is to say a tilted plan of angle $\ alpha$ ;

a tank is released without initial speed; to find its acceleration by application of the theorem of the kinetic energy.

tank is of total mass m +M, the caterpillar being of mass M; the distances between centres of the wheels has, the ray of the wheels is R.

Solution

The kinetic energy of translation is: 1/2 (m+M) V ² The energy of rotation of the caterpillar in the reference frame of the center of mass is 1/2 M (V/2) ².

One from of deduced the solution easily:

= G sin $\ alpha$ has. (m+M)/(m+M.2)

Demonstration

Indeed, each element of the caterpillar turns at the speed V in Rg; EC. total is thus worth: 1/2 (m +M.2) V ². As the caterpillar turns without slipping on the inclined level, the work of the force of the plan inclined on the plan is NULL. The conservation of energy thus gives: EC. + (m+M) gz = cste, from where the preceding result by derivation.

Note: the distances between centres of the wheels, has, does not play any part, nor the ray, R, of the wheels!

Let us reduce by the thought the mass of the frame, m, like negligible: 1/2 are found: relevant with the case = 0 have, because then it is about a rim descending a tilted plan (1/2Mr ² (V/r) ² +1/2MV ²).

See too

Mechanical of the Dynamic solid
of rotation

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