Question

A ring of radius R is placed such that it lies in a vertical plane. The ring is fixed, A bead of mass m is constrained to move along the ring without any friction.One end of the spring is connected with the mass m and other end is rigidly fixed with the topmost point of the ring. Initially the spring is in un-extended position and the bead is at a vertical distance R from the lowermost point of the ring.The bead is now released from rest.
a) What should be the value of spring constant K such that the bead is just able to reach bottom of the ring.
b) The tangential and centripetal accelerations of the bead at initial and bottommost position for the same value of spring constant k.

Answer 1

Natural length of the spring is $\sqrt{2}R$. When the bead just reaches the bottom of the ring, it has zero velocity and:

$\left(a\right)$ $kx=mg$

$⟹k(2R-\sqrt{2}R)=mg$

$⟹k=\frac{mg}{\sqrt{2}R(\sqrt{2}-1)}$

$⟹k=\frac{mg(\sqrt{2}+1)}{3\sqrt{2}R}$

$\left(b\right)$ Initially, when the bead is at $A$, there is no spring force acting on it. Therefore, centripetal acceleration initially $=$ zero.

Tangentially acceleration:

$\alpha =\frac{\tau }{I}$

At $A$, the torque is due to gravity:

$\therefore \alpha =\frac{mgR}{m{R}^2}$

$\therefore \alpha =\frac{g}{R}$

At $B$, the bead just reaches. Therefore, its centripetal acceleration is zero, and since at this point no tangential forces are acting on it, the tangential acceleration is also zero.