Question

At t = 2.55 s a point on the rim of a 0.230-m-radius wheel has a tangential speed of 53.5 m/s as the wheel slows down with a tangential acceleration of constant magnitude 12.0 m/s².

(a) Calculate the wheel's constant angular acceleration.
[-52 rad/s²]
(b) Calculate the angular velocities at t = 2.55 s and t = 0.

ω2.55 s	= [ ] rad/s
ω0	= [ ] rad/s

(c) Through what angle did the wheel turn between t = 0 and t = 2.55 s?
[ ] rad

(d) At what time will the radial acceleration equal g?
[ ]s after t = 2.55 s

Answer 1

a. angular accel in rad/s^2 = tangential accel / r = 10.7m/s^2 / 0.23m = 46.5 rad/s^2 b. angular velocity in rad/s = tangential speed / r = 45.0m/s / 0.23m = 196 rad/s c. accel is negative (slowing down) so velocity at t=0 = u = v + a*t = 196 + (46.5)*2.5 rad/s = 312 rad/s d. angle s = u*t - 0.5*a*t^2 s = (312*2.5) - 0.5*46.5*(2.5)^2 = 780 rad - 145 rad = 635 radians e. radial accel = v^2/r = 9.81m/s^2 v = sqrt(9.81*0.23) = 1.50 m/s = 1.50 / 0.23 rad/s = 6.53 rad/s This occurs when t = (u - v) / a = (312 - 6.53) / -46.5 = 6.57s