Reach for the Summit - P-S1-A5

Classical Mechanics Level 2

As shown above, A B , B C , C D , A D AB,BC,CD,AD are four rigid rods connected by hinges whose lengths are L L , and it's obvious that quadrilateral A B C D ABCD is a rhombus.

Initially, the diagonal B D BD is longer than A C AC , and the rhombus is put on the horizontal ground, then A , C A,C is moving to the opposite side along line A C AC with the same magnitude of velocity v v .

What's the acceleration ( m / s 2 ) (m/s^2) of point B B relative to the ground at the moment when the rhombus becomes a square (see the picture on the right)?

Submit the value when L = 2 m , v = 10 m / s L=\sqrt{2}\ m,\ v=10\ m/s .

Reach for the Summit problem set - Physics


The answer is 200.

Alice Smith by Alice Smith , 20, China

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1 solution

Karan Chatrath
Jul 16, 2020

Placing an inertial coordinate frame of reference for the system such that its origin is the centre of the rhombus and the X-axis points rightwards and Y axis points upwards. Let θ \theta be the angle the link B C BC makes with the horizontal.

X coordinate of C C is:

x c = L cos ⁡ θ x_c = L\cos{\theta} x ˙ c = v = − L θ ˙ sin ⁡ θ \dot{x}_c = v = -L\dot{\theta} \ \sin{\theta} at θ = π / 4 \theta = \pi/4 : θ ˙ = − v 2 L \dot{\theta} = -\frac{v\sqrt{2}}{L}

Now it is given that the velocity of point C C is a constant:

⟹ x ¨ c = 0 = − L θ ¨ sin ⁡ θ − L θ ˙ 2 cos ⁡ θ \implies \ddot{x}_c = 0 = -L\ddot{\theta} \ \sin{\theta} - L\dot{\theta}^2 \ \cos{\theta}

at θ = π / 4 \theta = \pi/4 : ⟹ θ ¨ = − θ ˙ 2 \implies \ddot{\theta} = -\dot{\theta}^2

Now, consider point B B

y b = L sin ⁡ θ y_b = L\sin{\theta}

Double differentiating:

y ¨ b = L θ ¨ cos ⁡ θ − L θ ˙ 2 sin ⁡ θ \ddot{y}_b = L\ddot{\theta} \ \cos{\theta} - L\dot{\theta}^2 \ \sin{\theta}

Plugging in values: θ = π / 4 \theta = \pi/4 and the expressions for θ ˙ \dot{\theta} and θ ¨ \ddot{\theta} gives the required answer of:

∣ y ¨ b ∣ = 200 \lvert \ddot{y}_b \rvert = 200

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Hello, the problem that wrongly reported is one that I finally got wrong. I thought of correcting my mistake and posting a solution anyway, but I realised later that I cannot post one. Moreover, I cannot find my mistake.

The answer I get is v m ≈ 2.20605 v_m \approx 2.20605 while the correct answer is something very close. Could you please share your solution?

Karan Chatrath - 11 months ago

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