Let's toss a coin!

Classical Mechanics Level 4

You are pitted against Lee Chong Wei in a superseries final!
You are given a chance to flip the coin for the toss.
You take a coin with heads facing the sky, and
toss it with an initial velocity v v and angular velocity ω \omega .
He claims tails on the toss.
He will definitely lose if _________ . \text{\_\_\_\_\_\_\_\_\_} .

Details and Assumptions:

  • Ignore air resistance.

  • n n is a positive integer.

  • The coin lands at the same level it was tossed from.

  • He only loses the toss!

If you are looking for more such twisted questions, Twisted problems for JEE aspirants is for you!
3 v ω = n π g 3v\omega = n\pi g 6 v ω = 5 n π g 6v\omega = 5n\pi g v ω = n π g v\omega = n\pi g 2 v ω = ( 2 n + 1 ) π g 2v\omega = (2n+1)\pi g

Rohith M.Athreya by Rohith M.Athreya , 22, India

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1 solution

The equation of vertical motion for the coin is y = y 0 + v t − 1 2 g t 2 y = y_{0} + vt - \frac{1}{2}gt^{2} .

The time the coin is in the air before it is back at its starting point is then given by

0 = v t − 1 2 g t 2 ⟹ t = 2 v g 0 = vt - \frac{1}{2}gt^{2} \Longrightarrow t = \dfrac{2v}{g} .

For LCW to lose, the coin must make n n full revolutions while in the air, ensuring that it lands heads up. We thus require that ω t = 2 n π \omega t = 2n\pi for some positive integer n n . Plugging in our value for t t gives us that

ω × 2 v g = 2 n π ⟹ v ω = n π g \omega \times \dfrac{2v}{g} = 2n\pi \Longrightarrow \boxed{v\omega = n\pi g} .

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