A 40 kg child is standing at the edge of a rotation merry-go-round. The merry-go

Q: A 40 kg child is standing at the edge of a rotation merry-go-round. The merry-go

given m = 40 kg d = 4 m radius, r = d/2 = 4/2 = 2 m w = 10 rpm = 10*2*pi/60 = 1.047 rad/s a) centripeatl acceleration, a = r*w^2 = 2*1.047^2 = 2.19 m/s^2 = 2.19*g/9.8 = 0.2235*g b) Force acting on the child = N*mue_s m*a = m*g*mue_s a = g*mue_s ==> mue_s = a/g = 2.19/9.8 = 0.2235

ID: 584236 • Letter: A

Question

A 40 kg child is standing at the edge of a rotation merry-go-round. The merry-go-round has a rotational speed of 10 rpm and has a diameter of 4 meters. Determine the centripetal accleration of the child. Express your answer in units of m/s^2 and also in g's (Recail 1 g-98 m/s^2) Using the five step process described in lecture determine the minimum coefficient of friction that is needed so that the child does not lose contact with the merr-go-round. Will time type of friction be static or kinetic?

Explanation / Answer

given

m = 40 kg

d = 4 m

radius, r = d/2 = 4/2 = 2 m

w = 10 rpm

= 10*2*pi/60

= 1.047 rad/s

a) centripeatl acceleration, a = r*w^2

= 2*1.047^2

= 2.19 m/s^2

= 2.19*g/9.8

= 0.2235*g

b) Force acting on the child = N*mue_s

m*a = m*g*mue_s

a = g*mue_s

==> mue_s = a/g

= 2.19/9.8

= 0.2235

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A 40 kg child is standing at the edge of a rotation merry-go-round. The merry-go

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