JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Work, Power & Energy

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A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of 4 ms⁻¹, is. ______ cm.
(take g=10 ms⁻²)

Answer: 120

Solution

When the cylinder unwinds and falls through height h, it gains both translational and rotational kinetic energy. By conservation of energy:

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

For a solid cylinder,

I = \frac{m R ^{2}}{2}

and

v = R ω

, so:

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} \cdot \frac{m R ^{2}}{2} \cdot \frac{v ^{2}}{R ^{2}} = \frac{1}{2} m v^{2} + \frac{1}{4} m v^{2} = \frac{3}{4} m v^{2}

g h = \frac{3 v ^{2}}{4}

h = \frac{3 \times 16}{4 \times 10} = \frac{48}{40} = 1.2 m = 120 cm

Answer: 120

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Work, Power & Energy

Solution

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