Work Power and Energy 4 Question 4
5. A stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed $u$. The magnitude of the change in its velocity as it reaches a position, where the string is horizontal, is
(a) $\sqrt{u^{2}-2 g L}$
(b) $\sqrt{2 g L}$
(c) $\sqrt{u^{2}-g L}$
(d) $\sqrt{2\left(u^{2}-g L\right)}$
(1998, 2M)
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Answer:
Correct Answer: 5. (d)
Solution:
- From energy conservation, $v^{2}=u^{2}-2 g L$
Now, since the two velocity vectors shown in figure are mutually perpendicular, hence the magnitude of change of velocity will be given by $|\Delta \mathbf{v}|=\sqrt{u^{2}+v^{2}}$
Substituting value of $v^{2}$ from Eq.(i), we get
$$ \text { 6. } \begin{aligned} |\Delta \mathbf{v}| & =\sqrt{u^{2}+u^{2}-2 g L}=\sqrt{2\left(u^{2}-g L\right)} \\ \text { or } \quad a _c & =k^{2} r t^{2} \\ \frac{v^{2}}{r} & =k^{2} r t^{2} \end{aligned} $$
or $\quad v=k r t$
Therefore, tangential acceleration, $a _t=\frac{d v}{d t}=k r$
or $\quad$ Tangential force, $F _t=m a _t=m k r$
Only tangential force does work.
$$ \text { Power }=F _t v=(m k r)(k r t) $$
$$ \text { or } \quad \text { Power }=m k^{2} r^{2} t $$
