Question

Laurel and Hardy are sitting on a merry-go-round while it turns at a steady speed. Laurel weighs 2/3 of what Hardy weighs. Hardy is sitting twice as far from the center as Laurel. Both Laurel and Hardy are just on the verge of slipping. Which statement below correctly describes the relation of their coefficients of friction? A) Laurel’s coefficient is twice as big as Hardy’s. B) Hardy’s coefficient is twice as big as Laurel’s.                       C) Laurel’s coefficient is 3 times as big as Hardy’s. D) Hardy’s coefficient is 6 times as big as Laurel’s. E) The coefficients are the same.

Answer 1

suppose steady speed =  ω

supose mass of  Hardy = M

r = 2d

mass of Laurel = 2/3 M

r = d

they are moving in a merry-go-round so the only frictional force will keep them seated , which balances the centrifugal force outward  

so

m*ω² *r = μ*mg

μ*g = ω² *r

μ = ω² *r/g

since  ω  and g are the same for both of them  

μ ∝ r

for  Hardy

μ = 2d

and for Laurel

μ = d

so  

Hardy’s coefficient is twice as big as Laurel’s.

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