Question

A bead of mass $10\text{g}$ oscillates along the horizontal diameter $AB$ of the ring of radius $R=5\text{cm}$. At an instant shown in the figure, kinetic energy of the particle is $0.015\text{J}$. Find the force applied $\left(\text{in Newtons}\right)$ by the ring on the bead.

• A. $0.33$
• B. $0.75$
• C. $0.66$
• D. $0.60$

Answer 1

The correct option is C $0.66$

Let $N$ be the normal reaction between the bead and the ring at given instant, then from $FBD$ of the bead


$N-mg\sin {37}^o=\frac{m{v}^2}{R}$
$⟹N=mg\sin {37}^o+\frac{2K}{R}$
where $K$ is kinetic energy of the bead.
$⟹N=0.01\times 10\times \frac{3}{5}+\frac{2\times 0.015}{5\times {10}^{-2}}$
$⟹N=\frac{0.3}{5}+\frac{3}{5}$
$⟹N=\frac{3.3}{5}=0.66\text{N}$
Hence option C is the correct answer