Question

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°?

Answer 1

Answer:

 ω = 0.467 rad/s

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

$I = \dfrac{1}{2} m R^2$

$I = \dfrac{1}{2}\times 144 \times 2.75^2$

    I = 544.5 kg.m²

torque = force  x radius

τ = 37.7 x  2.75

τ = 103.675 N.m

angular acceleration

$\alpha= \dfrac{\tau}{I}$

$\alpha= \dfrac{103.675}{544.5}$

 α = 0.190 rad/s²

now ,

distance = $33.2\times \dfrca{2\pi}{360}$

d = 0.579 rad

we know,

using equation of rotational motion

$d = \omega t + \dfrac{1}{2}\alpha t^2$

$0.579 = \dfrac{1}{2}\times 0.190\times t^2$

 t = 2.46 s

angular speed

 ω =  α  x t

 ω = 0.19 x 2.46

 ω = 0.467 rad/s