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

Respuesta :

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

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

F= 233.3N

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