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Question:

A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5m. 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 :-

0.02

0.5

0.3

0.28

Solution:

Correct option is C. 0.5
Angular impulse = change in angular momentum
τΔt = ΔL
mg(l/2) × 0.01 = m(l²/3)ω
ω = 3g × 0.01 / (2l) = 3 × 10 × 0.01 / (2 × 0.3) = 1/2 = 0.5 rad/s
time taken by rod to hit the ground
t = √(2h/g) = √(2 × 5/10) = 1 sec
in this time angle rotate by rod
θ = ωt = 0.5 × 1 = 0.5 radian