At t = 3.25 s a point on the rim of a 0.225-m-radius wheel has a tangential spee

Question

At t = 3.25 s a point on the rim of a 0.225-m-radius wheel has a tangential speed of 51.0 m/s as the wheel slows down with a tangential acceleration of constant magnitude 10.5 m/s2. (a) Calculate the wheel's constant angular acceleration. -46.67 Correct: Your answer is correct. rad/s2 (b) Calculate the angular velocities at t = 3.25 s and t = 0. 3.25 s = 204 Incorrect: Your answer is incorrect. rad/s 0 = 355.678 Incorrect: Your answer is incorrect. rad/s (c) Through what angle did the wheel turn between t = 0 and t = 3.25 s? 109.202 Incorrect: Your answer is incorrect. rad (d) At what time will the radial acceleration equal g? 7.48 Incorrect: Your answer is incorrect. s after t = 3.25 s

Explanation / Answer

a. angular accel in rad/s^2 = tangential accel / r = 10.5m/s^2 / 0.225m = 46.67 rad/s^2

b. angular velocity in rad/s = tangential speed / r = 51.0m/s / 0.225m = 226.67 rad/s

accel is negative (slowing down)
so velocity at t=0 = u = v + a*t = 226.67+ (46.67)*3.25 rad/s = 378.34 rad/s

c. angle s = u*t - 0.5*a*t^2
s = (378.34*3.25) - 0.5*46.67*(3.25)^2 = 983.14 radians

d. radial accel = v^2/r = 9.81m/s^2
v = sqrt(9.81*0.225) = 1.48 m/s = 1.48 / 0.225 rad/s = 6.6 rad/s
This occurs when t = (u - v) / a = (378.34 - 6.6) / 46.67 = 7.96 sns

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