Reach for the Summit - P-S1-A2

Classical Mechanics Level 2

As shown above, a rod with length l l is leaning against the vertical wall. The lowest point of the rod, A A is moving right at velocity v 0 v_0 , at this time, the angle of the rod and the ground is α \alpha .

Then there exists a point on the rod which has the minimum magnitude of velocity. Find the minimum velocity.

Take α = 30 ° \alpha=30 \degree , and the minimul velocity is λ v 0 \lambda v_0 . Submit ⌊ 10000 λ ⌋ \lfloor 10000 \lambda \rfloor .

Reach for the Summit problem set - Physics


The answer is 8660.

Alice Smith by Alice Smith , 20, China

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1 solution

Let the position vector of a point on the rod be r ⃗ = i ^ p l cos ⁡ α + j ^ ( 1 − p ) l sin ⁡ α \vec r=\hat i pl\cos α+\hat j (1-p)l\sin α

v 0 = d d t ( l cos ⁡ α ) = − l sin ⁡ α d α d t v_0=\dfrac {d}{dt}(l\cos α)=-l\sin α\dfrac {dα}{dt}

Magnitude of velocity of the point is

v = ∣ d r ⃗ d t ∣ = v 0 p 2 + ( 1 − p ) 2 cot ⁡ 2 α v=|\frac{d\vec r}{dt}|=v_0\sqrt {p^2+(1-p)^2\cot^2 α}

= v 0 4 p 2 − 6 p + 3 =v_0\sqrt {4p^2-6p+3}

= v 0 ( 2 p − 3 2 ) 2 + 3 4 ≥ 3 2 v 0 =v_0\sqrt {(2p-\frac{3}{2})^2+\frac{3}{4}}\geq \dfrac {\sqrt 3}{2}v_0

So, λ = 3 2 \lambda =\dfrac{\sqrt 3}{2} , and the answer is 8660 \boxed {8660} .

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