A uniform rod of mass M = 2 kg and length L is suspended by two smooth hinges 1 and 2 as…
Physics · JEE Advanced · NTA Exams — Rotational Motion
A uniform rod of mass M = 2 kg and length L is suspended
by two smooth hinges 1 and 2 as shown in the figure. A force
F = 4 N is applied downward at a distance L/4 from hinge 2 .
Due to the application of force F, hinge 2 breaks. At this
instant, applied force F is also removed. The rod starts to
rotate downward about hinge 1 .
The acceleration of the end point of the rod, when the rod becomes vertical is
- 30 m/s2
- 20 m/s2
- 10 m/s2
- 0
Answer
(A) 30 m/s 2
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