Laws of Motion 4 Question 5
5. A long horizontal rod has a bead which can slide along its length and is initially placed at a distance $L$ from one end $A$ of the rod. The rod is set in angular motion about $A$ with a constant angular acceleration $\alpha$. If the coefficient of friction between the rod and bead is $\mu$, and gravity is neglected, then the time after which the bead starts slipping is
$(2000,2 M)$
(a) $\sqrt{\frac{\mu}{\alpha}}$
(b) $\frac{\mu}{\sqrt{\alpha}}$
(c) $\frac{1}{\sqrt{\mu \alpha}}$
(d) infinitesimal
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Answer:
Correct Answer: 5. (a)
Solution:
- Tangential force $\left(F _t\right)$ of the bead will be given by the normal reaction $(N)$, while centripetal force $\left(F _c\right)$ is provided by friction $\left(f _r\right)$. The bead starts sliding when the centripetal force is just equal to the limiting friction.
$$ F _t \text { is inwards } $$
Therefore,
$$ F _t=m a=m \alpha L=N $$
$\therefore \quad$ Limiting value of friction
$$ \left(f _r\right) _{\max }=\mu N=\mu m \alpha L $$
Angular velocity at time $t$ is $\omega=\alpha t$
$\therefore$ Centripetal force at time $t$ will be
$$ F _c=m L \omega^{2}=m L \alpha^{2} t^{2} $$
Equating Eqs. (i) and (ii), we get
$$ t=\sqrt{\frac{\mu}{\alpha}} $$
For $t>\sqrt{\frac{\mu}{\alpha}}, F _c>\left(f _r\right) _{\max }$ i.e. the bead starts sliding.
In the figure, $F _t$ is perpendicular to the paper inwards.
