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Question

A uniform rod of mass m and length l is placed over a smooth horizontal surface along the y-axis and is at rest as shown in figure an impulsive force F is applied for a small time Δt along x-direction at point A. The x-coordinate of end A of the rod when the rod becomes parallel to x-axis for the first time is :

[Initially, the coordinate of centre of mass of the rod is (0, 0)]

120436_a9cbf3be97da411eb8be0e41214dc466.png

A
πl12
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B
l2(1+π12)
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C
l2(1π6)
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D
l2(1+π6)
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Solution

The correct option is B l2(1+π6)
As torque = change in angular momentum
FΔt=mv (linear momentum) ......(i)
and (Fl2)Δt=ml212ω (angular momentum).....(ii)
Dividing Eqs. (i) and (ii), we get
2=12vωlω=6vl
Using S=ut,
Displacement of CM is π2=ωt=(6vl)t
and x=vt
Dividing, we get
2xπ=l6x=πl12
Coordinates of A will be [πl12+l2,0].

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