Answer:
Explanation:
Given that the hoop makes one complete oscillation in 2s, this implies that the period of oscillation is 2s
Then,
T = 2s
Let Mass of the thin hoop be M
Let Radius of the hoop be R
Moment of inertial of a hoop is
I = MR²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
M is the mass of the hoop
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = R
Then, applying the formula
T = 2π √ (I / MgR)
So, we know that
I = MR²
Then,
T = 2π √( MR² / MgR)
T = 2π √(R/g)
Make R subject of formula
Square both sides
T² = 4π²(R/g)
T²g = 4π²R
Then, R = T²g / 4 π²
Since g = 9.8m/s² and T= 2s
R = 2² × 9.8 / 4π²
R = 0.993m
Then, the radius of the hoop is 0.993m
The radius of the hoop that makes one complete small-angle oscillation each 2.0 s is determined as 1.985 m.
The angular speed of the hoop after 2.0 seconds is calculated as follows;
ω = 2πN
ω = 2π(1/2)
ω = π rad/s = 3.142 rad/s
v = ωr
From principle of conservation of energy
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
v = √2gr
ωr = √2gr
(ωr)² = 2gr
ω²r² = 2gr
ω²r = 2g
r = 2g/ω²
r = (2 x 9.8)/(3.142)²
r = 1.985 m
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