Question

3. At t-0, angular velocity of a wheel is 24 rad/s, and its angular acceleration is 30 rad/s. (i) How much angle does the wheel turn through after 2 seconds of rotation? (ii) At t-2 sec., the wheel suddenly begins to undergo angular deceleration (negative angular acceleration) of 8.16 rad/s'. What additional time from t -2 secs, does it take the wheel to come to a stop? Hint: Use appropriate rotational kinematic equation to answer different parts of the question. (A) (i) 127 rad; (ii) 12.3 sec. (B) (i) 108 rad; (ii) 14.7 sec. (C) (i) 100 rad; (ii) 12.3 sec. (D) (i) 108 rad; (ii) 10.3 sec.

Answer 1

Given that: at t = 0 angular velocity w_0 = 24 rad/s angular acceleration alpha = 30 rad/s^2 at t = 2 sec. using theta = w_0 t + 1/2 alpha t^2 implies theta = 24 times 2 + 1/2 times 30 times (2)^2 implies theta = 48 + 60 implies theta = 108 radians so, wheel turn through 108 rad in 2 sec. after t = 2 sec, alpha' = -8.16 rad/s^2 using, w(t = 2 sec) = w_0 + alpha t at t = 2 sec, w(t = 2sec) = 24 + 30 times 2 = 84 rad/s now, again w = w(t = 2sec) + alpha' t implies 0 = 84 - 8.16 times t implies t = 84/8.16 = 10.294 sec hence, wheel will stop after 10.3 sec. so correct option is(D).