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 τ=0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in…

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 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

Accepted Answer

(1) 0.5. Two stages: a very short pivot on the edge that sets the box spinning, then a free fall at that steady spin. While it pivots about the edge, the weight acts at the centre, a distance l/2 away, and the moment of inertia about the edge is (ml²)/3. α=(mgl/2)/((ml²)/3)=(3g)/(2l)=(3(10))/(2(0.3))=50 rad s⁻². After the short time τ: ω=ατ=50(0.01)=0.5 rad s⁻¹. The angle turned during that 0.01 s is 1/2ατ²=0.0025 rad, which is negligible. Once clear of the edge no torque acts about the centre of mass, so ω stays at 0.5 rad s⁻¹. Fall time from 5 m: t=√(2h)/g=√(10)/(10)=1 s. Angle turned: θ=ω t=0.5 rad.

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