Kinematics Problem: Disc/Rod assembly - maximum speed

A disc is centered at the origin O ( 0 , 0 ) O(0,0) and is free to rotate about its center. A rod P A PA is attached to the disc at point P P and to another rod that can only move horizontally along the x x -axis, at point A A . Both P P and A A are revolute joints. If O P = 50 c m , P A = L = 150 c m \overline{OP} = 50 cm , \overline{PA} = L = 150 cm , and the disc rotates at a constant angular velocity of ω = d θ d t = 0.25 r a d / s e c \omega = \dfrac{d\theta}{dt} = 0.25 rad/sec , where θ \theta is the counter clockwise angle that O P OP makes with the positive x x -axis, find the maximum speed of point A A , (speed = v | v | , and v = d x d t v = \dfrac{dx}{dt} ) , in c m / s e c cm / sec , where x x is the x x -coordinate of point A ( x x is negative).

Note: The answer is a positive number.


The answer is 13.183.

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1 solution

Karan Chatrath
Sep 11, 2020

O P = R OP = R A P = L AP = L

Coordinates of P P are:

x p = R cos θ x_p = R\cos{\theta} y p = R sin θ y_p = R\sin{\theta}

Say the link AP makes an angle ϕ \phi with the positive X-axis. Coordinates of point A A are then:

x a = L cos ϕ + R cos θ x_a = -L\cos{\phi} + R\cos{\theta} y a = 0 y_a = 0

This gives rise to the constraint equation for the Y coordinate of point P P which is:

y p = R sin θ = L sin ϕ y_p = R\sin{\theta} = L\sin{\phi}

Eliminating ϕ \phi from the equation of x a x_a gives the following equation (Simplification left out):

x a = R cos θ L 2 R 2 sin 2 θ x_a = R\cos{\theta} - \sqrt{L^2 - R^2 \sin^2{\theta}}

Differentiating wrt. time gives:

v a = ( R sin θ + R 2 sin θ cos θ L 2 R 2 sin 2 θ ) ω v_a = \left(-R\sin{\theta} + \frac{R^2 \sin{\theta} \ \cos{\theta}}{\sqrt{L^2 - R^2 \sin^2{\theta}}}\right) \omega

The above equation is v a v_a as a function of θ \theta which varies uniformly with time. This is a standard unconstrained maximisation problem which I solved using a script of code. I swept across the range 0 θ 2 π 0 \le \theta \le 2\pi and stored the maximum value.

The problem can also be solved by differentiating wrt θ \theta and equating the result to zero, solving, and doing a second derivative test. I have not done so, as programming is convenient here.

The result is:

v a , m a x 13.183 v_{a,max} \approx 13.183

@Karan Chatrath very nice solution. Upvoted.

Talulah Riley - 9 months ago

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