A rectangles solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time $\tau=0.01$ s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Rotational Mechanics

A rectangles solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time

τ = 0.01

s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

(A)0.5✓ Correct
(B)0.6
(C)0.02
(D)0.28

Answer: (A)

Solution

As the box tips over the edge its weight acts at the centre, a distance

\frac{ℓ}{2}

from the edge, giving a torque

m g \frac{ℓ}{2}

for the short time

τ = 0.01 s

. The angular impulse equals the angular momentum gained,

m g \frac{ℓ}{2} τ = I ω

, and for a bar turning about its end

I = \frac{m ℓ ^{2}}{3}

. Solving,

ω = \frac{3 g τ}{2 ℓ} = \frac{3 \times 10 \times 0.01}{2 \times 0.3} = 0.5 rad/s

. After leaving the edge it falls

5 m

, taking

t = \frac{2 h}{g} = \frac{2 \times 5}{10} = 1 s

. In that time it turns through

θ = ω t = 0.5 \times 1 = 0.5 radian

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Rotational Mechanics

Solution

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