Statistics132

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

Particle on a string.zip (20.0 KB, 5 views)

skeeter

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

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

$\Delta x = \sqrt{3ga} \cdot \dfrac{1}{2} \cdot \sqrt{\dfrac{a}{g}} = a \cdot \dfrac{\sqrt{3}}{2}$

Statistics132

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.

skeeter

Had cosine on the brain, typed sine ... \$#!+ happens.