Lateral Acceleration and Angular Dynamics

A point-mass is connected to point P P by a mass-less rigid rod of length 2 m 2 \, \text{m} . Point P P accelerates to the right at ( a = 3 m/s 2 ) (a = 3 \, \text{m/s}^2) . There is an ambient downward gravitational acceleration of ( g = 10 m/s 2 ) (g = 10 \, \text{m/s}^2) .

The rod makes an angle θ \theta with the vertical. The variation in θ \theta over time can be described by the following differential equation:

θ ¨ = A c o s θ + B s i n θ \large{\ddot{\theta} = A \, cos \, \theta + B \, sin \, \theta }

In the above expression, A A and B B are real numbers with units of s 2 s^{-2} . What is the value of ( A + B ) (A + B) ?

Note: Assume that the coordinates for point P P are ( P x = 1 2 a t 2 ) (P_x = \frac{1}{2} a \, t^2) and ( P y = 0 ) (P_y = 0) , with the system starting into motion at t = 0 t = 0 . At t = 0 t = 0 , θ = 0 \theta = 0 .


The answer is -3.5.

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1 solution

Laszlo Mihaly
Jan 21, 2018

We select an accelerating frame of reference that moves together with point P. In this frame of reference there are two forces acting on the mass: a downward force of gravity m g mg and a horizontal "inertial" force of m a -ma , where the - sign indicates that this force points to the left. The equation of motion is m g l sin θ + m a l cos θ = m l 2 d 2 θ d t 2 -mgl \sin \theta+ mal \cos \theta = ml^2 \frac{d^2 \theta}{ dt^2} , where m l 2 ml^2 is the moment of inertia. This yields A = a / l = 1.5 s 2 A= a/l= 1.5 s^{-2} and B = a / l = 5.0 s 2 B= a/l= -5.0 s^{-2} . Therefore A + B = 3.5 A+B=-3.5 .

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