Question

23) A disc launcher, accelerates a disc from rest at a speed of 25.0m / s when turning it 1.35 rev. Suppose the disk moves in the arc of a circle of 1.00m radius. a) Calculate the final angular speed of the disk. b) Determine the magnitude of angular acceleration of the disk, assuming it is constant. c) Calculate the time interval required for the disk to accelerate from rest to 25.0m / s.

Answer 1

initial speed=0

final speed=25 m/s

angle covered=1.35 revolution=1.35*2*pi

=8.4823 rad

radius of the circle=1 m

then final angular speed=linear speed/radius=25 rad/s

part a:

final angular speed =25 rad/s

part b:

initial angular speed=0

final angular speed=25 rad/s

angle turned=8.4823 rad

then using the formula:

final angular speed^2-initial angular speed^2=2*angular acceleration*angle turned

==>25^2-0^2=2*angular acceleraion*8.4823

==>angular acceleration=36.841 rad/s^2

part c:

time taken=(final angular speed-initial angular speed)/angular acceleration

=0.67858 seconds