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A constant torque of `31.4 N-m` id exterted on a pivoted wheel. If the angular acceleration of the wheel is `4 pi rad//s^2`, then the moment of inertia will be.A. `2.5 kg-m^(2)`B. `3.5 kg-m^(2)`C. `4.5 kg-m^(2)`D. `5.5 kg-m^(2)` |
Answer» `tau=I alphaimplies I=(tau)/(alpha)=(3.14)/(4pi)=2.5kg m^(2)` | |