InterviewSolution
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InterviewSolution
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The angular position of a point on the rim of a rotating wheel is given by `theta=4t^(3)-2t^(2)+5t+3` rad. Find (a) the angular velocity at `t=1 s`, (b) the angular acceleration at `t=2 s`. (c ) the average angular velocity in time interval `t=0` to `t=2 s` and (d) the average angular acceleration in time interval `t=1` to `t=3 s`. |
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Answer» `theta=4t^(3)-2t^(2)+5t+3` `omega=(d theta)/(dt)=12t^(2)-4t+5` `alpha=(domega)/(dt)=24t-4` (a) At `t=1 s`, `omega=12(1)^(2)-4(1)+5=13 rad//s` (b) At `t=2 s`, `alpha=24(2)-4=44 rad//s^(2)` (c ) `t_(1)=0, theta_(1)=3 rad` `t_(2)=2 s, theta_(2)=4(2)^(3)-2(2)^(2)+5(2)+3=37 rad` `undersetomega(-)=(theta_(2)-theta_(1))/(t_(2)-t_(1))=(37-3)/(2-0)=17 rad//s` (d) At `t_(1)=1 s`, ` omega_(1)=12(1)^(2)-4(1)+5=13 rad//s` At `t_(2)=3 s`, `omega_(2)=12(3)^(2)-4(3)+5=101 rad//s` `undersetalpha(-)=(omega_(2)-omega_(1))/(t_(2)-t_(1))=(101-13)/(3-1)=44 rad//s^(2)` |
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