Question

A string is wrapped several times around the rim of a small hoop. The hoop has radius 0.08 m and mass 1.2 kg. If the free end of the string is held in place and the hoop is released from rest, calculate a) the tension in the string while the hoop is descending; 5.88 N b) the time it takes the hoop to descend 0.5 m; 0.452 s c) the angular velocity of the rotating hoop after it has descended 0.5 m

Answer 1

Hello From your question you already have the answers for ( a ) and ( b ) hence I will provide the solution to option ( c )

answer :

a) 5.88 N

b) 0.452 s

c) 27.68 rad/s

Step-by-step explanation:

Hoop radius ( R ) = 0.08 m

mass of hoop = 1.2 kg

height hoop descended ( h ) = 0.5 m

Calculate The angular velocity of the rotating hoop after it descends 0.5 m

since it is released from rest we will apply the law of conservation

mgh = 1/2 mv^2 + 1/2 * Ic * ω^2 ----- ( 1 )

where : Ic (moment of inertia ) = m*R^2 , ω = angular velocity , v = Rω

Back to equation 1 above

mgh = m*R^2*ω^2 ------- ( 2 )

therefore angular velocity ( ω ) =
`\sqrt{(gh)/(R^2) }` , h = 0.5 m, g = 9.81, R = 0.08 m

∴ ω = 27.68 rad/s