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Homework Statement
A massless stick is pinned at point A. And it has a concentrated ball mass of 1kg attached to it at point B.
Given the initial values:
Initial angular position θ=45◦
Initial angular velocity ω=0
Initial angular acceleration α=0
Show that because of the setup of the problem, the concentrated mass B follows a circular path of:
[tex]x = \sqrt{2}cos\theta[/tex][tex]y = \sqrt{2}sin\theta[/tex]
Homework Equations
The tangential acceleration is induced by the gravitational force on the mass:
[tex]ma_t = mgcos\theta[/tex]
The stick reacts to the compressive force induced by the ball mass, which is also the centripetal acceleration:
[tex]ma_c = AB = mgsin\theta[/tex]
The Attempt at a Solution
Find the forces acting at point B:
[tex]\sum F_x= ma_x[/tex][tex]AB_x = ma_x[/tex][tex]ABcos\theta = ma_x[/tex][tex]mgsin\theta cos\theta = ma_x[/tex][tex]a_x = gsin\theta cos\theta[/tex]
[tex]\sum F_y= ma_y[/tex][tex]AB_y - mg = -ma_y[/tex][tex]-ABsin\theta + mg = ma_y[/tex][tex]m(g - gsin^2\theta) = ma_y[/tex][tex]a_y = g(1 - sin^2\theta)[/tex][tex]a_y = gcos^2\theta[/tex]
To find the x and y positions simply double integrate the respective accelerations ax and ay with respect to time.
[tex]x = \int \int a_x = \int \int gsin\theta cos\theta dt^2[/tex][tex]y = \int \int a_y = \int \int gcos^2\theta dt^2[/tex]
However, the problem is that θ is a function of time or θ(t). In addition, although there is a constant vertical gravitational acceleration, the angular acceleration is not constant. Therefore, the equation for constant angular acceleration is not applicable:
[tex]\theta = \frac{1}{r}(\theta _0 + \omega _0 t + \frac{1}{2}\alpha _0 t^2)[/tex]
So, in order to find θ, i have to use the tangential acceleration relation with the angular acceleration:
[tex]r\alpha = a_t[/tex][tex]\alpha = \frac{a_t}{r}[/tex][tex]r\theta = \int \int \frac{a_t}{r}dt^2[/tex]
However, because the tangential acceleration is known in terms of θ, it sets me back to my original problem that i am trying to integrate θ with respect to time:
[tex]\theta = \frac{1}{r^2}\int \int gcos\theta dt^2[/tex]