JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Rotational Mechanics

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A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60° with vertical ? [g is the acceleration due to gravity]

(A)The radial acceleration of the rod's center of mass will be $\frac{3 g}{4}$ ✓ Correct
(B)The angular acceleration of the rod will be $\frac{2 g}{L}$
(C)The angular speed of the rod will be $\frac{3 g}{2 L}$ ✓ Correct
(D)The normal reaction force from the floor on the rod will be $\frac{Mg}{16}$ ✓ Correct

Answer: (A), (C), (D)

Solution

Using work-energy theorem from vertical to 60°:

m g \frac{ℓ}{4} = \frac{1}{2} (\frac{m ℓ ^{2}}{3}) ω^{2}

Solving:

ω = \frac{3 g}{2 ℓ}

. Option (C) is correct.

Radial acceleration of CM:

a_{r} = \frac{ℓ}{2} ω^{2} = \frac{3 g}{4}

. Option (A) is correct.

Using torque about contact point:

\frac{m g ℓ}{2} sin 60° = \frac{m ℓ ^{2}}{3} α

Solving:

α = \frac{3 3 g}{4 ℓ}

, not

\frac{2 g}{L}

. Option (B) is incorrect.

Net vertical acceleration:

a_{v} = a_{r} cos 60° + a_{t} cos 30° = \frac{15 g}{16}

.

Applying Newton's second law vertically:

m g - N = m (\frac{15 g}{16})

Solving:

N = \frac{m g}{16}

. Option (D) is correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Rotational Mechanics

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