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Moment of inertia of a body about a given axis is the rotational inertia of the body about that axis. It is respresented by `I = MK^(2)`, where `M` is mass of body and `K` is radius of gyration of the body about that axis. It is a scalar quantity, which is measured in `kg m^(2)`. when a body rotates about a given axis, and teh axis of rotates also moves, then total `K.E.` of body `= K.E.` of translation `+ K.E.` of rotation `E = (1)/(2)mv^(2) + (1)/(2)I omega^(2)` Which the help of the compreshension given above, choose the most apporpriate altermative for each of the following questions : Kinetic energy of rotation of the flywheel in the above case isA. `20 J`B. `20 J`C. `395 J`D. `80 J` |
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Answer» Correct Answer - C `K.E.` of rotation `= (1)/(2)I omega^(2) = (1)/(2)I (2pin)^(2)` `= (1)/(2) xx 5 (2pi xx(120)/(60))^(2) = 395.1 J` |
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