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A person exerts a tangential force of 37.7 N on the rim of a disk-shaped merry-go-round of radius 2.75 m and mass 144 kg. If the merry-go-round starts at rest, what is its angular speed after the person has rotated it through an angle of 33.2°?

User Proseidon
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1 Answer

4 votes

Answer:

ω = 0.467 rad/s

Step-by-step explanation:

given,

tangential force exerted by the person = 37.7 N

radius of merry-go-round = 2.75 m

mass of merry-go-round = 144 Kg

angle = 33.2°

moment of inertia


I = (1)/(2) m R^2


I = (1)/(2)* 144 * 2.75^2

I = 544.5 kg.m²

torque = force x radius

τ = 37.7 x 2.75

τ = 103.675 N.m

angular acceleration


\alpha= (\tau)/(I)


\alpha= (103.675)/(544.5)

α = 0.190 rad/s²

now ,

distance =
33.2* \dfrca{2\pi}{360}

d = 0.579 rad

we know,

using equation of rotational motion


d = \omega t + (1)/(2)\alpha t^2


0.579 = (1)/(2)* 0.190* t^2

t = 2.46 s

angular speed

ω = α x t

ω = 0.19 x 2.46

ω = 0.467 rad/s

User Joaoal
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