Question

A particle of mass $m$ is attached to inner portion of periphery of a ring of radius $R = 1$ m and mass $m$. System is slightly disturbed when the particle is at the highest point of the ring. Normal reaction between the ring and the ground is N and angular velocity of system is $\omega$ rad/s. (Given $m = 1$ kg, $g = 10$ m/s$^2$). There is no slipping between the ring and the ground at any instant. The value of N (in newton) when the particle is at lowest position of the ring is ...........

Answer 1

20 + ω^2

Answer 2

Explanation of the solution:

1. For the particle: When it is at the lowest point (inside the ring), its circular motion (radius 1 m) requires a centripetal force upward equal to
     $m\omega^2$.
2. The weight of the particle is $mg$ (downward). Thus the contact (normal) force from the ring on the particle, acting upward, must provide the net force:   $N_p = mg + m\omega^2 \,.$
3. By Newton’s third law, the particle exerts an equal and opposite force on the ring, i.e. a downward force of magnitude $N_p = mg + m\omega^2$.
4. Now, the ring (mass $m$) also has its own weight $mg$ acting downward. The only upward force on the ring is the normal reaction $N$ from the ground. Since the ring’s centre does not move vertically (no slip against the ground), we have equilibrium:   $N = mg \;(\text{ring}) + \bigl(mg + m\omega^2\bigr) \,.$
5. Substituting $m = 1$ kg and $g = 10$ m/s$^2$:   $N = 10 + (10 + \omega^2) = 20 + \omega^2 \quad \text{(in newtons)}\,.$

Answer:
$N = 20 + \omega^2$ newtons.