A rod CD of length L and mass M is placed horizontally on a frictionless horizontal surface as shown. A second identical rod AB which is also placed horizontally (perpendicular to CD) on the same hori

Question

A rod CD of length L and mass M is placed horizontally on a frictionless horizontal surface as shown. A second identical rod AB which is also placed horizontally (perpendicular to CD) on the same horizontal surface is moving along the surface with a velocity v in a direction perpendicular to rod CD and its end B strikes the rod CD at end C and sticks to it rigidly. Then:

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Answer:

B, C

Solution:

$M v = 2 M v^{1} \to v^{1} = \frac{v}{2} 2 (\frac{M L ^{2}}{12} + M (\frac{L}{2 2})^{2}) ω = M v \frac{L}{4}$

TestHub · Accepted Answer

​ M v = 2 M v 1 → v 1 = 2 v ​ ​ 2 ( 12 M L 2 ​ + M ( 2 2 ​ L ​ ) 2 ) ω = M v 4 L ​ ​

Physics - Rotation Question with Solution | TestHub

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