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A rod of mass `M = 5kg` and length `L = 1.5 m` is held vertical on a table as shown. A gentle push is given to it and it starts falling. Friction is large enough to prevent end A from slipping on the table. (a) Find the sum of linear momentum of all the particles of the rod when it rotates through an angle `theta = 37^(@)` (b) Find the friction force and normal reaction force by the table on the rod, when `theta = 37^(@)` (c) Find value of angle q when the friction force becomes zero. `[tan 37^(@) = (3)/(4) " and " g = 10 m//s]` |
Answer» Correct Answer - (a) `7.5 kg m//s` (b) `f = 9 N, N = 24.5 N` (c) `cos^(-1) ((2)/(3))` |
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