InterviewSolution
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InterviewSolution
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Average torque on a projectile of mass `m` (initial speed `u` and angle of projection `theta`) between initial and final positions `P` and `Q` as shown in figure, about the point of projection is : .A. `1(m u^2 sin 2 theta)/(2)`B. `(m u^2 cos theta)/(2)`C. `m u^2 sin theta`D. `m u^2 cos theta` |
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Answer» Correct Answer - A (a) `vec tau. Delta t = Delta vec L`…(i) Here `Delta t = time of flight = (2 u sin theta)/(g)` Change in angular about point of projection (initially it is zero) `|vec(Delta L)| = |vec L_f = vec L_i|= (m u sin theta)` Range =`((m u sin theta)(u^2 sin 2 theta))/(g) = (m u^2 sin theta sin 2 theta)/(g)` Now `|vec tau_(av)| = |vec(Delta L)/(Delta t)| = (m u^3 sin theta sin 2 theta)/(g) xx (g)/(2u sin theta)` =`(m u^2 sin 2 theta)/(2)`. |
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