Question

A heavy particle is suspended by a $1.5m$ long string. It is given a horizontal velocity of $\sqrt{57}m/s$. (a) Find the angle made by the string with the upward vertical, when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take $g=10ms{/}^2$.

Answer 1

Utillising the work-energy theorem, we first check if the particle completes a full circle. This happens when the tension in the string at the topmost point just approaches zero.

Initial Energy = $\frac{1}{2}m{u}^2=\frac{1}{2}m(\sqrt{57}{)}^2=23.5m$, (initially, particle has just kinetic energy)

Final Energy = $mgh=m\left(10\right)(2\times 1.5)=30m$, (purely potential in nature, since the velocity of the particle is zero at the topmost point if tension is to be zero)

Here u, g, h are constants as given in the question with their usual meanings, and m is the mass of the particle.

We see that initial energy is not enough to make the particle reach the topmost point. It is, however, sufficient to make the particle reach the upper half of the circle (i.e. $>15m$) Thus, the string will slacken somewhere in the upper half.

Assume that the string makes an angle of $\theta$ with the upward vertical just at the point of slackening, and is moving with a velocity of $v$ in the circular path. Thus, balancing forces along the radial direction, by taking components, we get:

$T+mg\cos \theta =\frac{m{v}^2}{r}$,

where $T$ is the tension in the string, and $r$ is the length of the rope.

At the point of slackening, $T=0$. Thus, we get:

$v^2=gr\cos \theta$

Applying conservation of energy between the initial state and this state:

$\frac{1}{2}m(\sqrt{57}{)}^2=\frac{1}{2}m{v}^2+mgh$

(here, $h$ refers to the height reached when the string slackens)

$\Rightarrow \frac{1}{2}m(\sqrt{57}{)}^2=\frac{1}{2}m{v}^2+mg(r+r\cos \theta )$

Replacing $v^2=gr\cos \theta$, and dividing throughout by $m$:

$\Rightarrow \frac{1}{2}(\sqrt{57}{)}^2=\frac{1}{2}gr\cos \theta +g(r+r\cos \theta )$

$\Rightarrow \frac{3}{2}rg\cos \theta =\frac{1}{2}(\sqrt{57}{)}^2-rg$

Putting in the values of $r$ & $g$, we get:

$\cos \theta =\frac{27}{45}$

$\therefore \theta ={\cos }^{-1}\left(\frac{27}{45}\right)={\cos }^{-1}\left(\frac{3}{5}\right)\to \text{(A)}$

Thus, the value of $v$:

$v=\sqrt{rg\cos \theta }=\sqrt{\left(1.5\right)\left(10\right)\left(\frac{3}{5}\right)}$

$\therefore v=3m/s\to \text{(B)}$

And the maximum height reached over the starting point:

$h=\frac{v^{2}si{n}^2\theta }{2g}$

$\Rightarrow h=\frac{3^{2}si{n}^2{53}^o}{2g}$

$\therefore h=0.16m$

$Totalheight=(0.16+0.9)m=1.06m\to \text{(C)}$