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A grindstone increases in angular speed from `4.00 rad//s` to `12.00 rad//s` in `4.00 s`. Through what angle does it turn during that time interval if he angular acceleration is constant ?A. 8.00 radB. 12.0 radC. 16.0 radD. 32.0 rad |
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Answer» Correct Answer - D (d) The angular displacement will be `Delta theta = omega_(avg) Delta t = ((omega_f + omega_i)/(2)) Delta t` =`((12.00 rad//s + 4.00 rad//s)/(2))(4.00 s) = 32.0 rad`. |
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