The angular position of a point on the rim of a rotating whe

The angular position of a point on the rim of a rotating wheel is given by = 6.73t - 6.52t^2 + 2.74t^3, where is in radians and t is in seconds. What are the angular velocities at (a) t = 3.81 s and (b) t = 7.37 s? (c) What is the average angular acceleration for the time interval that begins at t = 3.81 s and ends at t = 7.37 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

Solution

w = dtheta/dt = 6.73 - 13.04t + 8.22t^2
a) w(3.81) = 6.73 - 13.04(3.81) + 8.22(3.81)^2 = 119.322 rad/s

b)w(7.37) = 6.73 - 13.04(7.37) + 8.22(7.37)^2 = 357.110 rad/s

c) ave alpha = delta w/ delta t = (357.110-119.322)/2 = 118.89 rad/s^2

d) a = dw/dt = -13.04+16.44t

atbeginning t=0 so a= -13.04 rad/s^2

e) at t=7.37 a= -13.04 +16.44*7.37 = 108.112 rad/s^2