A child stands at the rim of a rotating merry-go-round of radius 2.6 m . The chi

Question

A child stands at the rim of a rotating merry-go-round of radius 2.6 m. The child has a mass of 46.0kg, and the carousel turns with an angular velocity of 2.8 rad/s. What is the child's centripetal acceleration?

What is the horizontal force required between her feet and the floor of the carousel to keep her in the circular path?

What minimum coefficient of static friction is required? (This will tell you whether she is likely to be able to stay on the merry-go-round.)

Explanation / Answer

here r = 2.6 m m = 46 Kg w = 2.8 rad/s

(a) as we kow that the centripetal acceleration

ac = V2 / r

here V = wr so

ac = w2r

ac = 2.8 x 2.8 x 2.6 = 20.384 m/s2   Ans

(b) the force rrequired for keep the circular motion is called as centripetel force

so as we know that

F = mV2 / r = mw2r

F = 46 x 2.6 x 2.8 x 2.8 = 937.664 N Ans

(c) aswe know that

F = usN

us = F / mg

= 937.664 / (46 x 9.8)

= 2.08 Ans

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