Answer:
Vcm = 2.71m/s
Explanation:
From the problem description, the upper end is held fixed and not pulled upward so no work is done on the system of the hoop and string by holding this end of the string.
Mechanical energy is conserved because the string does not slip, although there is friction between the string and hoop.
So we can use the equation of conservation of mechanical energy.
K1 + U1 = K2 + U2
K1, K2 = initial and final kinetic energies of the system
U1, U2 = initial and final potential energies
Vcm = velocity of the center of mass of the system.
R = radius of the hoop
K1 = 0 system was initially at rest
U1 = mgh
K2 = 1/2MVcm² + 1/2×I×ω²
I = moment of inertia of the loop
I = MR²
ω = Vcm/R = Angukar frequency
K2 = 1/2×MVcm² + 1/2× MR² ×(Vcm/R)²
K2 = 1/2MVcm² + 1/2MVcm²
K2 = MVcm²
U2 = 0 chosen reference point as the system falls a distance of h
h = 75cm = 0.75m
Substituting the respective terms in the mechanical energy conservation equation above
0 + Mgh = MVcm² + 0
Mgh = MVcm²
gh = Vcm²
Vcm = √gh = √9.8×0.75 = 2.71m/s.
