A 50.0 kg child stands at the rim of a merry-go-round ofradius 1.65 m, rotating

Question

A 50.0 kg child stands at the rim of a merry-go-round ofradius 1.65 m, rotating withan angular speed of 2.50 rad/s. (a) What is the child's centripetal acceleration?
m/s2

(b) What is the minimum force between her feet and the floor of thecarousel that is required to keep her in the circularpath?
N

(c) What minimum coefficient of static friction isrequired?
(a) What is the child's centripetal acceleration?
m/s2

(b) What is the minimum force between her feet and the floor of thecarousel that is required to keep her in the circularpath?
N

(c) What minimum coefficient of static friction isrequired?

Explanation / Answer

   Mass of the child, M =50 kg
   Radius, r = 1.65 m
    Angular velocity, = 2.5 rad / s
(a)
    Centripetal acceleration, ac =2 r
                                            = 3.12 * 1.65
                                            = 10.3125 m /s^2
(b)
    Minimum force required, F min = Centripetalforce
                                                   = M ac
                                                                    = 50 * 10.3125
                                                   = 515.625 N
(c)
    Frictional force = Required centripetalforce
            M g = 515.625
    Coefficient of friction, = 515.625 / ( 50 *9.8 )
                                        = 1.05229

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