A thin uniform rod of length 3m and mass M is held horizontally by two vertical strings attached to the two ends, as shown in figure. One of the strings is cut. Find the angular acceleration of the rod at the moment it is cut. (Take g=10m/s2).
A
15rad/s2
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B
10rad/s2
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C
5rad/s2
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D
7.5rad/s2
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Solution
The correct option is C5rad/s2 Given, length of rod l=3m Just after cutting the string at B, the rod will tend to rotate about its other end i.e A, which is attached with a string.
Moment of inertia of rod about about end A, I=Ml23...(i)
From the FBD of rod, since T passes through point A, hence τT=0 For Mg the perpendicular distance from A is r⊥=l2 ∴τMg=Mg(l2)...(ii)
Net external torque on rod about A is τnet=τT+τMg=τMg Applying the fundamental equation for torque: τnet=Iα...(iii)
From Eq (i).(ii),(iii) Mg×(l2)=(Ml23)α or α=32gl Substituting l=3m, ∴α=3×102×3=5rad/s2