A 165-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim

Question

A 165-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s? (State the magnitude of the force.) N

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niczorrrr 5 years 2021-07-24T03:39:20+00:00 1 Answers 14 views 0

Answers ( )

  1. sori
    0
    2021-07-24T03:41:03+00:00
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    Answer:

    233.3N

    Explanation:

    Given:

    radius ‘r’= 1.5m

    mass ‘m’= 165kg

    time ‘t’=2 sec

    angular speed ‘ω’= 0.6 rev/s

    the magnitude of a torque is given by:

    τ = F . r = I . α

    where, ‘α’ is the angular acceleration and  ‘I’ is the rotational inertia

    F=  I . α/ r =>[ (\frac{1}{2} . m . r^{2})(2π . ω/t) ]/r

    F=[( \frac{1}{2} . 165 . 1.5^{2})(2π . 0.6/2)]/1.5

    F= 233.3N

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