Question

A bead of mass $m$ is attached to one end of a spring of natural length $R$ and spring constant $k=\frac{mg}{R}.$ The other end of the spring is fixed at point $A$ on a smooth vertical ring of radius $R$ as shown in the figure. The tangential acceleration of the bead just after it is released will be:

• A. $13.4g{\text{cm/s}}^2$
• B. $0.134g{\text{cm/s}}^2$
• C. $13.4g{\text{m/s}}^2$
• D. Can't be determined

Answer 1

The correct option is A $13.4g{\text{cm/s}}^2$

From the geometry shown in figure, it is evident that $AB\ne R$ and there is an extension in the spring , let's say ${}^{\prime }{x}^{\prime }$.


$x=$ length of $AB$ - natural length of spring
$\therefore x=(R\cos {30}^{\circ }+R\cos {30}^{\circ })-R$
$=2R\cos {30}^{\circ }-R=(\sqrt{3}-1)R$
Now, spring force:
$\left(F\right)=kx=\frac{mg}{R}(\sqrt{3}-1)R=(\sqrt{3}-1)mg$

From FBD:

Applying equation of circular motion along tangential direction:
$mg\sin {30}^{\circ }-F\sin {30}^{\circ }=m\times a_t$
$\Rightarrow m{a}_t=\frac{mg}{2}-\frac{F}{2}$
$\Rightarrow m{a}_t=\frac{mg}{2}-\frac{(\sqrt{3}-1)mg}{2}$
$\Rightarrow m{a}_t=\frac{2mg-\sqrt{3}mg}{2}$
$\Rightarrow a_t=\frac{(2-\sqrt{3})g}{2}$

$\therefore$ $|a_t|=0.134g{\text{m/s}}^2=13.4g{\text{cm/s}}^2$

If actual value of $a_t$ comes negative, then it represents that actual direction of $a_t$ is opposite of what we assumed in FBD.