JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Laws of Motion

A spring mass system (mass m, spring constant k and natural length

ℓ

) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about its axis with an angular velocity

ω

(k ≫ m ω^{2})

the relative change in the length of the spring is best given by the option:

(A) $\frac{m ω ^{2}}{3 k}$
(B) $\frac{m ω ^{2}}{k}$ ✓ Correct
(C) $\frac{2 m ω ^{2}}{3 k}$
(D) $\frac{2}{3} (\frac{m ω ^{2}}{k})$

Answer: (B)

Solution

As the disc spins, the block needs an inward centripetal force

m ω^{2} (ℓ + x)

, supplied by the stretched spring's pull

k x

. Balancing these,

k x = m ω^{2} (ℓ + x)

, so

x = \frac{m ω ^{2} ℓ}{k - m ω ^{2}}

. Because

k ≫ m ω^{2}

, the denominator is essentially

k

, and the fractional change in length is

\frac{x}{ℓ} \approx \frac{m ω ^{2}}{k}

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Laws of Motion

Solution

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