Question

A block is tied to one end of a light string of length

l

whose other end is fixed to a rigid support. The block is given a speed of

\sqrt{7 g l / 2}

from the lowermost position. Find the height and speed at which the block leaves the circle. Also find the maximum height to which it rises finally.

Accepted Answer

As velocity of the block $v_{b}, \sqrt{2 g l} < v_{b} < \sqrt{5 g l} .$

The block will leave the circle at some point $P,$ where the radius $O P$ makes an angle $θ$ with the upward vertical.

From $A$ to $P,$

$Δ K + Δ U = 0$ or $Δ K = - Δ U$

Loss in $KE =$ Gain in GPE

$\frac{1}{2} m v_{b}^{2} - \frac{1}{2} m v^{2} = m g (l + l \cos θ) \dots (i)$

From the force diagram: $T + m g \cos θ = \frac{m v^{2}}{l}$
As the block leaves the circle at $P, T = 0 .$

$m g \cos θ = \frac{m v^{2}}{l} \dots (ii)$

Putting the value of $v^{2}$ from equation $(ii)$ in equation $(i)$ , we get,

$\frac{1}{2} m v_{b}^{2} - \frac{1}{2} m (g l \cos θ) = m g l (1 + \cos θ)$

$\Rightarrow \frac{1}{2} m (\frac{7}{2} g l) = \frac{m g l}{2} (2 + 3 \cos θ)$
$\Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = 60 °$

Hence, the block leaves the circle and string slack at a height $h = l + l \cos 60 ° = 1.5 l$ from the bottom. The velocity at that moment $v$ is $v = \sqrt{lgcos θ} = \sqrt{\frac{l g}{2}}$ .

After the string becomes slack, the block moves as a projectile in parabolic path. Now, further height attained

$= \frac{v^{2} \sin^{2} θ}{2 g} = \frac{\lg}{4 g} (\frac{3}{4}) = (\frac{3}{16}) l$

Thus, total height attained from the bottom is,

$\frac{3}{2} l + \frac{3}{16} l = \frac{27}{16} l$

B M Sharma Solutions for Chapter: Circular Motion, Exercise 51: ILLUSTRATION 9.48

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 51: ILLUSTRATION 9.48

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 51: ILLUSTRATION 9.48 with Hints & Solutions

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