Question

A string is wrapped several times around the rim of a small hoop with a radius of 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest.
After the hoop has descended 75.0 cm, calculate (a) the angular speed of the rotating hoop and (b) the speed of its center.

Answer 1

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.