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November 11th, 2016, 02:28 PM  #1 
Member Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1  A particle of mass m attached to a light inextensible string of the length a.
One end of a light inextensible string of length $a$ is attached to a particle of mass $m$. The other end is attached to a fixed point $O$ which is at a height $\frac{5}{2} a$ above the horizontal ground. Initially the string is taut and horizontal. The particle is then projected vertically downward with velocity $\sqrt {2ag}$. When the string has turned through an angle $θ$ ($<π$) the velocity $v$ is $\sqrt {2ag(1+\sinθ)}$ and the tension, $T$, in the string is $mg(2 + 3\sinθ)$. The string can withstand a tension of $5mg$, giving $5mg = 2mg + 3mg\sinθ$, so $θ = \arcsin(1) = \frac{π}{2}$ from the horizontal position, so the string does not break. If the string can withstand a tension of at most $\frac{7}{2}mg$, find the values of $θ$ and $v$ when the string breaks. In this case find the time to reach the vertical through $O$. Hence show that the particle strikes the ground at the point vertically below $O$. See the enclosed doc with a sketch and some workings so far. Last edited by skipjack; November 12th, 2016 at 02:56 PM. 
November 11th, 2016, 03:46 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387  Quote:
$mg(2+3\sin{\theta}) \ge \dfrac{7mg}{2} \implies \sin{\theta} \ge \dfrac{1}{2} \implies \theta \ge \dfrac{\pi}{6}$ At $\theta = \dfrac{\pi}{6}$, $v_0 = \sqrt{3ga}$ with direction $\phi = \dfrac{\pi}{3}$, $h = \dfrac{5a}{2}  a\sin\left(\dfrac{\pi}{6}\right) = 2a$, and $x = a\sin\left(\dfrac{\pi}{6}\right) = a \cdot \dfrac{\sqrt{3}}{2}$ $\Delta y = v_0 \sin{\phi} \cdot t  \dfrac{1}{2}gt^2$ $0 = 2a  \sqrt{3ga} \cdot \dfrac{\sqrt{3}}{2} \cdot t  \dfrac{1}{2}gt^2$ quadratic formula yields $t = \sqrt{\dfrac{a}{g}}$ $\Delta x = v_0 \cos{\phi} \cdot t$ $\Delta x = \sqrt{3ga} \cdot \dfrac{1}{2} \cdot \sqrt{\dfrac{a}{g}} = a \cdot \dfrac{\sqrt{3}}{2}$  
November 12th, 2016, 01:14 PM  #3 
Member Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 
In the third line down, it should say $x=a\cos(\frac{π}{6}) = a\cdot\frac{√3}{2}$ or could use $x=a\sin(\frac{π}{3}) = a\cdot\frac{√3}{2}$. Thanks for the response. Last edited by skipjack; November 12th, 2016 at 02:49 PM. 
November 12th, 2016, 01:46 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 
Had cosine on the brain, typed sine ... \$#!+ happens.


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attached, inextensible, length, light, mass, particle, string 
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