When you stand on the merry-go-round in question 12, you hold a string from which
is suspended a rubber stopper of mass 45 g. You are 2.9 m from the centre of the
merry-go-round. You take 4.1 s to complete one revolution. What angle does the string make with the vertical?

Respuesta :

The angle the string makes with the vertical is : 34.6°

Given data :

mass of rubber stopper ( m ) = 45 g

Radius ( r ) = 2.9 m

Time to complete one revolution ( T ) = 4.1 secs

Calculate the angle the string makes with the vertical

applying the formula below

Tan ∅ = [tex]\frac{mw^2r}{mg}[/tex] --- ( 1 )

cancel out m from equation ( 1  )

Tan ∅ = [tex]\frac{w^2r}{g}[/tex]    ----- ( 2 )

where ; w = [tex]\frac{2\pi }{T}[/tex]  

Therefore equation ( 2 ) becomes

Tan ∅ = [tex]\frac{4\pi ^2r}{T^2g}[/tex]   --- ( 3 )

where ; r = 2.9 m, T = 4.1 secs , g = 9.8

Insert values into equation ( 3 )

Tan ∅ = 0.69

      ∅ = Tan ( 0.69 )

         = 34.6°

Hence we can conclude that The angle the string makes with the vertical is : 34.6°.

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