Solution: Bead on a Ring (Full Derivation)
Consider a ring of radius \( r \) centered at the origin \( (0,0) \). The two nails are fixed at points \( A \) and \( B \) on the diameter. Let the nails be symmetric about the center at distance \( a = 0.5r \). Coordinates: \( A(-0.5r, 0) \) and \( B(0.5r, 0) \). The bead is at a general point \( P \) on the ring: \( P(r \cos \theta, r \sin \theta) \).
We need to find the total length squared of the two strings \( l_1^2 + l_2^2 \), where \( l_1 = AP \) and \( l_2 = BP \).
Using the distance formula squared:
$$ l_1^2 = (x_P – x_A)^2 + (y_P – y_A)^2 = (r \cos \theta + 0.5r)^2 + (r \sin \theta)^2 $$ $$ l_2^2 = (x_P – x_B)^2 + (y_P – y_B)^2 = (r \cos \theta – 0.5r)^2 + (r \sin \theta)^2 $$Expanding the terms:
$$ l_1^2 = r^2 \cos^2 \theta + r^2 \cos \theta + 0.25r^2 + r^2 \sin^2 \theta $$Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
$$ l_1^2 = r^2 + 0.25r^2 + r^2 \cos \theta = 1.25r^2 + r^2 \cos \theta $$Similarly for \( l_2^2 \):
$$ l_2^2 = r^2 \cos^2 \theta – r^2 \cos \theta + 0.25r^2 + r^2 \sin^2 \theta $$ $$ l_2^2 = 1.25r^2 – r^2 \cos \theta $$Adding them together:
$$ l_1^2 + l_2^2 = (1.25r^2 + r^2 \cos \theta) + (1.25r^2 – r^2 \cos \theta) $$ $$ l_1^2 + l_2^2 = 2.5r^2 = \text{constant} $$Option (d): Total Potential Energy \( U = \frac{1}{2}k(l_1^2 + l_2^2) = \frac{1}{2}k(2.5r^2) = 1.25kr^2 \). This is constant.
Option (a): Since Potential Energy \( U \) is constant everywhere on the ring, the force gradient \( F = -\nabla U = 0 \). There is no tangential force to accelerate or decelerate the bead. Thus, it moves with its initial speed constant.
Option (c): Since the speed is constant and radius is constant, the magnitude of angular momentum \( L = mvr \) is conserved about the center O.
Conclusion: Options (a), (b), (c), and (d) are all correct.