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November 11th, 2016, 03:28 PM   #1
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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.
Attached Files
File Type: zip Particle on a string.zip (20.0 KB, 5 views)

Last edited by skipjack; November 12th, 2016 at 03:56 PM.
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November 11th, 2016, 04:46 PM   #2
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Quote:
If the string can withstand a tension of at most $\dfrac{7mg}{2}$, 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.
string will break when $T \ge \dfrac{7mg}{2}$ ...

$mg(2+3\sin{\theta}) \ge \dfrac{7mg}{2} \implies \sin{\theta} \ge \dfrac{1}{2} \implies \theta \ge \dfrac{\pi}{6}$

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 $t = \sqrt{\dfrac{a}{g}}$

$\Delta x = v_0 \cos{\phi} \cdot t$

$\Delta x = \sqrt{3ga} \cdot \dfrac{1}{2} \cdot \sqrt{\dfrac{a}{g}} = a \cdot \dfrac{\sqrt{3}}{2}$
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November 12th, 2016, 02:14 PM   #3
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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 03:49 PM.
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November 12th, 2016, 02:46 PM   #4
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Had cosine on the brain, typed sine ... \$#!+ happens.
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