A coin placed on a rotating turntable just slips if it is placed at a distance of 16cm from the center. If the angular velocity of the turntable is doubled, it will just slip at a distance of-

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1 cm

2 cm

4cm

8 cm

answer is C.

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Detailed Solution

Concept: Since the coin is rotating, there will be a centripetal force working on it in the outwards direction. Let that centripetal force be equal to

m ω^{2} r

( formula of centripetal force in terms of omega) . If the angular haste at which the coin is rotating is ω.
As it's given in the question that the coin is slipping, there will be a frictional force working on it. Let that frictional force be f.
The force working down on the coin will be mg which will be equal and contrary to the force working overhead on it which is N.
The formula of disunion is f = μN
According to the question, the coin slips. So, for the slipping to do, the frictional force will be equal to f = mω2r( related to the figure as well) .
The below formula can be said as equation 2.
⇒ f = mω2r( equation 2)
Observing equation 1 and equation 2, we can say that-
⇒ μN = mω2r
As we formerly know, N is equal to mg- N = mg. So, putting this value in the below equation we get-
⇒ μN = m ω2 ⇒ μ mg = m ω2r ⇒ μg =ω2r
Let this be equation 3.
Since the coin is placed at 16 cm from the center, the compass will be calculated as 16 cm.
The original angular haste of the coin was ω.
The final angular haste will be ω ′.
Since we doubled the angular haste, we can say that-
⇒ ω ′ = 2ω
The original compass as r. Let the final compass be r ’.
Comparing the original and the final conditions through equation 3, we can say that-
⇒ ω2r = ω ′ 2r ′
Putting the values of ω ′ and r, we get-
⇒ ω2 × 16 = ( 2ω) 2 × r ′ ⇒ ω2 × 16 = 4ω2 × r ′ ⇒ 16 = 4r ′ ⇒ r ′ = 164 ⇒ r ′ = 4 cm
Hence, the correct option is 3.

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