Physics Angular accelerations?

The angular position of a point on the rim of a rotating wheel is given by θ = 9.76t - 1.90t2 + 2.35t3, where θ is in radians and t is in seconds. What are the angular velocities at (a) t = 2.62 s and (b) t = 8.70 s? (c) What is the average angular acceleration for the time interval that begins at t = 2.62 s and ends at t = 8.70 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

1 Answer
Apr 12, 2018

Angular position is given as

#θ = θ = 9.76t - 1.90t^2 + 2.35t^3#

Angular velocity #omega# is given by

#omega(t)=(d theta)/dt=d/dt( 9.76t - 1.90t^2 + 2.35t^3)#
#=>omega(t)= 9.76 - 3.80t + 7.05t^2# .......(1)

(a) #omega(2.62)= 9.76 - 3.80xx2.62 + 7.05xx(2.62)^2#

#omega(2.62)=48.19802\ rad cdot s^-1#

(b) #omega(8.70)= 9.76 - 3.80xx8.70 + 7.05xx(8.70)^2#

#omega(2.62)=510.3145\ rad cdot s^-1#

(c) Average angular acceleration #alpha_"ave"=(Deltaomega)/(Delta t)#

#alpha_"ave"=(omega(8.70)-omega(2.72))/(8.70-2.72)#
#alpha_"ave"=(510.3145-48.19802)/(8.70-2.72)#
#alpha_"ave"=77.277\ rad cdot s^-2#

from (1) #alpha(t)=(domega(t))/dt=d/dt(9.76 - 3.80t + 7.05t^2)#

#=>alpha(t)=- 3.80 + 14.1t#

(d) Instantaneous angular accelerations at the beginning

#alpha(2.72)=- 3.80 + 14.1xx2.72#
#alpha(2.72)=34.552\ rad cdot s^-2#

(d) Instantaneous angular accelerations at the end

#alpha(8.70)=- 3.80 + 14.1xx8.70#
#alpha(2.72)=118.87\ rad cdot s^-2#