First slide

Vertical circular motion

Question

A stone of mass 1 kg tied to a light inextensible string of length L = $\frac{10}{3}$ m, whirling in a circular path in a vertical plane. The ratio of maximum tension to the minimum tension in the string is 4. If g is taken to be 10 $m s^{- 2}$ , the speed of the stone at the highest point of the circle is

Moderate

A

10 $m s^{- 1}$
B

$5 \sqrt{2}$ $m s^{- 1}$
C

10 $\sqrt{3}$ $m s^{- 1}$
D

20 $m s^{- 1}$

Solution

Minimum tension is at topmost point (speed = v) and
maximum tension at bottommost point (speed = u).
$∵ \frac{T_{\max}}{T_{\min}} = \frac{m g + \frac{m u^{2}}{L}}{- m g + \frac{m v^{2}}{L}} = 4$
$\Rightarrow u^{2} = 4 v^{2} - 5 g L$ …(i)
and $u^{2} = v^{2} + 2 g (2 L)$ …(ii)
On solving Eqs. (i) and (ii), we get
$v = 10 {ms}^{- 1}$

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