Question

Chapter 5, Problem 5/025 (video solution to similar problem attached) Incorrect 0 at time t 2.7 s. 0. The disk has angular velocity a = 0.23 rad/s at t-0 and then it experiences an angular acceleration = 2.7e where e is in radians and o is in radians per second squar Point A of the circular disk is at the angular position Determine the angular position of point A at time 24 Answers: 6- [4.36 the tolerance is +/-4% rad Question Attempts: 4 of 5 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

Theta(0) = 0 and w(0) xi w_0 = 0.23 rad s^-1 at t = 0 angular acceleration, alpha = 2.7 theta where, theta is in rad. Alpha is in rad s^-2. Alpha = d^2 theta/d t^2 = dw/dt d theta/d theta = dw/d theta w (where, d theta/dt = w) 2.7 theta d theta = dw w w^2/2 = 2.7 theta^2/2 + c (where, c rightarrow integration constant) at t = 0, theta = 0, w = 0.23 (0.23)^2/2 = 0 + c implies c = 0.02645 tildetilde 0.03 w^2 = 2.7 theta^2 + 0.06 implies w = squareroot 2.7 theta^2 + 0.06 Now, w = d theta/dt implies d theta/squareroot 2.7 theta^2 + 0.06 = dt integral 1/squareroot 2.7 d theta/squareroot theta^2 + (0.15)^2 tildeequal integral dt ln[theta + squareroot theta^2 + (0.15)^2] + ln c tildeequal 1.64 t c rightarrow integration constant. but theta(0) = 0 ln(0.15) + ln c