Rotating Rod Acts As A Ball Launcher

Let's work on finding the velocity $v$ as a function of $\theta$ instead of $\text{s}$ since $\dfrac{p}{q}\text{ m}/\text{s}=\dfrac{p}{q}\text{ m}/\theta$ because $\omega = 1^{\circ}/\text{s}$ . In order to do that, we can find the distance $d$ of the very bottom of the ball from the pivot point, then take the derivative. So if we can find $d$ as a function of $\theta$ , then we are good.

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Let us instead define $\theta$ as the angle measure between the vertical line passing through the pivot point and the rod; this way, we start with $\theta=0^{\circ}$ and end with $\theta=90^{\circ}$ . Also, let's use radians to make things simpler. Thus, the angle the rod makes to the ground is $\dfrac{\pi}{2}-\theta$ .

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We now draw the auxiliary lines as shown. Note that $\varphi = \dfrac{\dfrac{\pi}{2}-\theta}{2} = \dfrac{\pi}{4}-\dfrac{\theta}{2}$ . Since $d=\dfrac{1}{\tan\varphi}=\cot\varphi$ , we have $d(\theta)=\cot\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)$

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Since $v(\theta)=d'(\theta)$ , all we need to do is find the derivative of $d(\theta)$ . This is a simple use of the chain rule: $\begin{aligned}v(\theta) &=\dfrac{\text{d}}{\text{d}x}\cot\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\\ &= \dfrac{\text{d}}{\text{d}\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\cot\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\cdot \dfrac{\text{d}}{\text{d}x}\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\\ &= -\csc^2\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)\cdot -\dfrac{1}{2}\\ &=\boxed{ \dfrac{1}{2}\csc^2\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)}\end{aligned}$

Now we need to find the certain $\theta$ where the end of the rod is tangent to the circle.

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Note that in the diagram, $\varphi= \tan^{-1}\left(\dfrac{1}{10}\right)$ . Since $\theta=\dfrac{\pi}{2}-2\varphi$ , we have that $\theta=\dfrac{\pi}{2}-2\tan^{-1}\left(\dfrac{1}{10}\right)$ .

Plugging this back into $v(\theta)= \dfrac{1}{2}\csc^2\left(\dfrac{\pi}{4}-\dfrac{\theta}{2}\right)$ , we have that $\begin{aligned}v\left(\dfrac{\pi}{2}-2\tan^{-1}\left(\dfrac{1}{10}\right)\right)&= \dfrac{1}{2}\csc^2\left(\tan^{-1}\left(\dfrac{1}{10}\right)\right)\\ &=\dfrac{1}{2}(\sqrt{101})^2\\ &=\dfrac{101}{2} \end{aligned}$ So our answer is $101+2=\boxed{103}$ .

It might be surprising that after just $10$ meters, the ball is already flying at a speed of $50.5\text{ m}/\text{s}$ .

Its quite simple.Make use of the fact that their velocities along a common normal must be exactly the same.This yields $wl=vsin2 \gamma$ where tan $\gamma$ =1/10..