only to merry-go rounds Short Answer Questions (10 Points Each) 7,) A 25.0kg chi

Question

only to merry-go rounds Short Answer Questions (10 Points Each) 7,) A 25.0kg child is sitting 1.5m from the center of a merry go round which is moving with an angular 7) A 25.0kg child is si frequency of rad/s (A) what isthe period of the merry-go around? (B.) what is the centripetal force required to hold the child in place( Tra mthemerry8o around? (B.)What is the centripetal force edtoholdthechi din place -26 x (0.025)2 red's N B) Cond o 625 7 .) A 250 N force pulls an object at along a frictionless surface at 0.25 m/s, (A.) what is the mass of the object? The object then a reaches a surface with friction, and the acceleration with friction is 0.15 m/s?. (B.) What is the coefficient of kinetic friction?

Explanation / Answer

The merry-go around is moving with an angular frequency of /60 rad/s, and has a child of 25 kg weight sitting 1.5 metres away from the centre.

Part A.) The period of motion for a body moving with angular frequecny of w is given as T = 2 / w

That is, T = 2 x 3.14 x 60 / = 120 seconds

Part B.) Here, the centripetal force would be given as: F = Mv2/R

Now, v = wR = ( / 60)*1.5 = 0.0785 m/s

That is, F = 25 x 0.0252 / 1.5 = 0.1027 N

NOTE: By T = ? he is pointing to the time period of the merry-go around. The approach as can been seen with the question is correct, however, there is a slight mistake in calculation of w which is equal to / 60 = 0.052 and doesn't actually come out to be 0.0166 rad/s

That would also mean that centripetal force calculated will also change.

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