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Sagot :
1.57 rad/s is the angular velocity of the hoop about its center of mass.
Solution:
Given that
Weight of hoop m = 1.2 kg
radius of hoop R = 65 cm = 0.65 m
Force = 7 N
Torque on the hoop Τ = F×R →equation (1)
Where F is applied force on hoop
We also know that
Τ = Iα[tex]R^{2}[/tex]
Where α is the angular acceleration
I is the moment of inertia of hoop
For hoop I = m[tex]R^{2}[/tex]
T = (m[tex]R^{2}[/tex])α →equation (2)
From equation 1 and 2
FR = (m)α
F = mRα
α = F/mR
α = 7/1.2×0.65 = 3.79 rad/s×s
α= 3.79 rad/s×s
We have to calculate distance covered by center of hoop in 5 s
d = (1/2)at×t →equation(3)
Relation between linear acceleration and radius is
a = αR
a = 3.79×0.65
a = 2.46 m/s×s
now from equation (3)
d = 0.5×2.46×4×4
d = 19.68 m
so the distance covered by center of hoop in 4 s is 19.68 m.
Part(2)
angular velocity of hoop
angular momentum about the center of the hoop
w = v/R
ωt = 2×π
t is given 4 s
ωt = 6.28
ω×4= 6.28
ω = 1.57 rad/s
Hence, 1.57 rad/s is the angular velocity of the hoop about its center of mass.
Learn more about angular velocity here:- https://brainly.com/question/6860269
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