A 50.0 kg childstands at the rim of a merry-go-round of radius 1.65 m, rotating

Question

A 50.0 kg childstands at the rim of a merry-go-round of radius 1.65 m, rotating with an angular speedof 2.50 rad/s. (a) What is thechild'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 thechild'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

(a)    the childs centripetal acceleration is givenby    ac = r 2        = (1.65 m) (2.50 rad/ s)2        = ........ m/ s2 (b)    for this we use the newtons second law ofmotion    the minimum force is    Fc = m ac         = (50.0 kg)(ac)         = ......... N (c)    according to the problem we can see that thecentripetal acceleration is produced by the force of    friction so the static frictional forceis    Fc = fs         = .....N    normal force is    n = m g       = (50.0 kg) (9.80 m /s2)       = ........ N    so the coeffiecient of friction is     = fs / n       = ........         = (50.0 kg)(ac)         = ......... N (c)    according to the problem we can see that thecentripetal acceleration is produced by the force of    friction so the static frictional forceis    Fc = fs         = .....N    normal force is    n = m g       = (50.0 kg) (9.80 m /s2)       = ........ N    so the coeffiecient of friction is     = fs / n       = ........

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