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asked in NEET by (1.9k points)
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A uniform  thin rod of weight W is supported  horizontal  by two vertical props at its ends   . At t=0, one of these supports is kicked  out . The force on the other support  immediately  there after  is??

Ans- W/4

1 Answer

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answered by (61.3k points)

Let the center of mass of the rod have downward acceleration a. Then

W-F=ma ---------(1)

The angular momentum equation implies

WL/2 = Iω, ----------(2)

where I=moment of inertia about an end = 1/3mL2.

One has ω=a/(L/2),

a relation that is true for small ω, i.e short times.

From (1) and (2), 


commented by (1.9k points)
How u had eq force × dis = i w,???   If I am not wrong . The equation of 2 must be of torque  i.e w× l/2 = I alpha .  Through  this we find the value of alpha than finally  acc and the put the value of a in eq 1 . Then the answer  will come w/4.

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