Question

A bead of mass $m=300gm$ moves in gravity free region along a smooth fixed ring of radius $R=2m$. The bead is attached to a spring having natural length $R$ and spring constant $k=10N/m$. The other end of spring is connected to a fixed point O on the ring. AB $=\frac{6R}{5}$. Line OB is diameter of ring.

• A. If normal reaction on bead due to ring at point A is zero, then speed of bead at A is $12m/s$.
• B. The rate of change in speed at this instant is $24m/s^2$
• C. If normal reaction on bead due to ring at point A is zero, then speed of bead at A is $8m/s$.
• D. The rate of change in speed at this instant is $18m/s^2$

Answer 1

Answer: (B), (C)

The correct options are
B The rate of change in speed at this instant is $24m/s^2$
C If normal reaction on bead due to ring at point A is zero, then speed of bead at A is $8m/s$.


The length of the spring, when bead is at A, is

$OA=\sqrt{(OB{)}^2-(AB{)}^2}=\sqrt{(2R{)}^2-{\left(\frac{6R}{5}\right)}^2}=\frac{8R}{5}$

Elongation in the spring = OA - natural length

$=\frac{8R}{5}-R=\frac{3R}{5}$

Also $cos\theta =\frac{\left(\frac{8R}{5}\right)}{2R}=\frac{4}{5}$ and $sin\theta =\left(\frac{\frac{6R}{5}}{2R}\right)=\frac{3}{5}$

Radial component of spring force $=kxcos\theta =\frac{m{v}^2}{R}$

$\Rightarrow k\left(\frac{3R}{5}\right)\left(\frac{4}{5}\right)=\frac{m{v}^2}{R}$

$\Rightarrow v=\sqrt{\left(\frac{12k{R}^2}{25m}\right)}=8m/s$

Tangential component of force $=kxsin\theta =k\left(\frac{3R}{5}\right)\left(\frac{3}{5}\right)=\frac{9kR}{25}$

$\Rightarrow m\frac{dv}{dt}=\frac{9kR}{25}$

Rate of change in speed $\frac{dv}{dt}=\frac{9kR}{25m}=24m/s^2$