A Stone's Throw

Classical Mechanics Level 5

A stone of mass m m is thrown straight upward from ground level ( y = 0 ) (y = 0) at speed v 0 v_0 , under the influence of ambient gravitational acceleration g g . Let the vertical coordinate of the stone be y ( t ) y(t) , and let t f t_f be the time at which the stone lands again on the ground. The kinetic energy, potential energy, and action for the path are:

E ( t ) = 1 2 m ( y ˙ 2 ( t ) ) U ( t ) = m g y ( t ) S = ∫ 0 t f ( E ( t ) − U ( t ) ) d t E(t) = \frac{1}{2} m \Big(\dot{y}^2 (t) \Big) \\ U(t) = m g \, y(t) \\ S = \int_0^{t_f} \Big(E(t) - U(t) \Big) \, dt

Consider a non-physical deformed trajectory, which preserves the starting and ending points of the path, both in space and time.

y ′ ( t ) = α y ( t ) y'(t) = \alpha \, y(t)

Let S ′ S' be the action for the deformed trajectory, keeping in mind that the final time t f t_f does not depend on α \alpha . Suppose we make a graph of d S ′ d α \large{\frac{d S'}{d \alpha}} vs. α \alpha , with d S ′ d α \large{\frac{d S'}{d \alpha}} on the vertical axis and α \alpha on the horizontal axis. As it turns out, this plot is a straight line.

If α 0 \alpha_0 is the value of α \alpha at which d S ′ d α = 0 \large{\frac{d S'}{d \alpha} = 0} , and M M is the slope of the graph, give your answer as the product of α 0 \alpha_0 and M M . What is the significance of α 0 \alpha_0 ?

Note: This problem is fairly easy to do by hand

Details and Assumptions:

  • m = 1 m = 1
  • v 0 = 10 v_0 = 10
  • g = 10 g = 10


The answer is 66.67.

Steven Chase by Steven Chase , 37, USA

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1 solution

We have y ′ = α ( v 0 t − 1 2 g t 2 ) , d y ′ d t = α ( v 0 − g t ) , t f = 2 v 0 g , E ( t ) = 1 2 m α 2 ( v 0 − g t ) 2 , U ( t ) = m g α ( v 0 t − 1 2 g t 2 ) y'=α(v_0t-\dfrac{1}{2}gt^2), \dfrac{dy'}{dt}=α(v_0-gt), t_f=\dfrac{2v_0}{g},E(t)=\dfrac{1}{2}mα^2(v_0-gt)^2, U(t)=mgα(v_0t-\dfrac{1}{2}gt^2) . Substituting all these, we get S ′ = m α 2 v 0 3 3 g − 2 m α v 0 3 3 g S'=\dfrac{mα^2v_0^3}{3g}-\dfrac{2mαv_0^3}{3g} , and d S ′ d α = 2 m v 0 3 3 g ( α − 1 ) \dfrac{dS'}{dα}=\dfrac{2mv_0^3}{3g}(α-1) . So d S ′ d α = 0 \dfrac{dS'}{dα}=0 at α = α 0 = 1 α=α_0=1 and the slope of the graph is M = 2 m v 0 3 3 g M=\dfrac{2mv_0^3}{3g} , such that α 0 × M = 2 m v 0 3 3 g α_0\times M=\dfrac{2mv_0^3}{3g} . Substituting values, we get α 0 × M = 200 3 = 66.67 α_0\times M=\dfrac{200}{3}=\boxed {66.67} . The condition α 0 = 1 α_0=1 makes the action S ′ S' an optimum. Since at α 0 = 1 , d 2 S ′ d α 2 = 2 m v 0 3 3 g α_0=1, \dfrac{d^2S'}{dα^2}=\dfrac{2mv_0^3}{3g} is positive, this value of α 0 α_0 corresponds to the minimum of S ′ S' , that is, in the path given by y = v 0 t − 1 2 g t 2 y=v_0t-\dfrac{1}{2}gt^2 , the action S ′ = S S'=S is the least , the minimum value of S ′ S' being − m v 0 3 3 g = − 100 3 ≈ − 33.33 -\dfrac{mv_0^3}{3g}=-\dfrac{100}{3}\approx {-33.33}

4 Helpful 2 Interesting 1 Brilliant 0 Confused

And, of course, the fact that α 0 = 1 \alpha_0=1 reflects the Principle of Least Action.

Mark Hennings - 1 year, 3 months ago

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Yeah, that's right. I forgot to mention that.

A Former Brilliant Member - 1 year, 3 months ago

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