Question

A bead of mass m is attached to one end of a spring of spring constant k. The other end of the spring is attached to a point on the perimeter of a smooth horizontal ring of radius R. The bead can slide on the ring and is released from the situation shown. Natural length of spring is 2R. Contact force between bead and ring at this moment is x mg. Find x² .

Answer 1

At release, the bead experiences gravitational force downward (mg) and spring force (F_s) acting towards the point of attachment on the ring. The contact force (N) between the bead and the ring provides the necessary centripetal force.

The vertical equilibrium implies:
N - mg = 0
So, N = mg (this establishes the relationship for gravitational force).

The spring is stretched from its natural length (2R) when the bead is at distance R from the center of the ring. Therefore, the spring's stretch is:
x_s = (R - 2R) = -R
The spring force, using Hooke’s Law (F_s = k * x_s), becomes:
F_s = -k(R - 2R) = kR (as F_s acts upwards).

Substituting k = (4mg/R) from the given data, we have:
F_s = (4mg/R) * R = 4mg.

The spring force equivalently acts against the contact force providing some centripetal input:
Centripetal force required = m * (v²/R) = N + F_s = mg + 4mg = 5mg.

From the centripetal force:
mv²/R = 5mg.
Dividing throughout by m and rearranging gives:
v²/R = 5g or v² = 5gR.

Using the relationship between contact force (N = xmg), we had N as mg. Setting them equal gives x = 5.

Calculate x²:
x² = 5² = 25.