I am struggling with the following problem. It is a physics problem, but the step I am stuck at is essentially a trigonometry question.
A particle $P$ rotates on the circumference of a wheel of radius $R$ with the center at height $H$ above the ground. The wheel rotates some angle $\theta$ from its original position, at which point the particle is flung off the wheel with some velocity $v_p$. The problem asks what the maximum height the particle will attain above the ground. To solve for this, I need the initial height of the particle as a function of $\theta$.
To solve for the initial height, I realized that the initial height of the particle was
$H-R$. The height it gains can then be modeled by some $H(\theta)$, where $H(0) = 0$ and $H(90°) = R$. From there, I got stuck. I wasn't sure if at $\theta = 45°$ the particle is at height $\frac{R}{2}$ or not, and I wasn't sure how I could relate the two with a general formula.
Any hints would be greatly appreciated.
Best Answer
Draw the perpendicular from $P$ to the vertical. Then in the right angle triangle formed, you get $$x=R\sin\theta$$ and the distance from the center (below the center) is $$y=R\cos\theta$$ Then the height above ground is $$h=H-y=H-R\cos\theta$$