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 = = 0.01 s, remaining essentially
horizontal. The angle by which it would rotate when it hits the ground will be in radians) close to :​

Answer 1

Explanation:

Given 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 5 m. When released, it slips off the table in a very short time = = 0.01 s, remaining essentially  horizontal. The angle by which it would rotate when it hits the ground will be in radians) close to  

• We know that Angular impulse = change in angular momentum.
• So τ x Δt
• = F x d x Δt
• So mg x l/2 x Δt = Iω – 0    (l/2 is perpendicular distance, also initial angular momentum = 0)
•   Now I moment of inertia end of rod = ml^2 / 3
• So mg l/2 Δt = ml^2 / 3 ω
• 3 g Δt / 2l = ω
• 3 x 10 x 0.01 / 2 x 0.3 = ω
• = 0.5 rad / sec
• We know that from Newton’s law of equation we get
• S = ut + 1/2 gt^2
• Here u = 0
• S =  1/2 gt^2
• Or t = √2S / g
• Or t = √2h / g
•       = √2 x 5 / 10
• So t = 1 sec is the time taken by the rod to hit the ground.
• So the angle rotated by the rod will be θ = ωt
•                                                                      = 0.5 rad/ sec x 1 sec

                                                                       = 0.5 radians