Mechanics Problem (2 Viewers)

ivanradoszyce · Thursday at 1:06 PM

I came across this problem from the Cambridge Ex 2 book, the Mechanics section.

Part (d) has me scratching my head. Obviously $\alpha$ cannot be 1. This results in an improper integral.

From part (c)

$\begin{align*} \dot{y}^2 &= \frac{3g(3 - 2y)}{3 + 2y} \,\,\, \text{which simplifies to } \\ \frac{dt}{dy} &= - \frac{1}{\sqrt{3g}} \sqrt{\frac{1 + \frac{2}{3}y}{1 - \frac{2}{3}y}} \,\,\, \text{where - indicates a decreasing rate } \end{align*}$

I can't determine what the $\alpha$ terminal would be if I performed the integration. I thought the upper
terminal could be 3/2 before applying a substitution of $u = \frac{2}{3} y$ . However that doesn't achieve the desired result. So I'm confused. The answer indicates $\alpha= \frac{1}{4}$

Any help appreciated.

tywebb · Thursday at 2:51 PM

ivanradoszyce · Thursday at 2:57 PM

Thanks tywebb, that's exactly what I got. I appears the answer at the end of the text is incorrect. Frustrating to say the least.

Mechanics Problem (2 Viewers)

ivanradoszyce

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tywebb

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ivanradoszyce

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