The shaft of a pump starts from rest and has an angular acceleration of 3.30 rad/s^2 for 16.0 s.?

Question

(a) At the end of this interval, what is the shaft's angular speed?

(b) What is the angle through which the shaft has turned?

Answer 1

To solve this problem you need two equations. For part a) you use the equation Final Angular Speed = Initial Angular Speed + Angular Acceleration * Time. For part b) you will use the equation Final Angular Position = Initial Angular Position + Initial Angular Speed * Time + 1/2 * Angular Acceleration * Time ^2.

For part a) you know the Initial Angular Speed = 0, Angular Acceleration = 3.30 rad/sec, and Time = 16.0 sec. So solving the equation for part a) gives:
Final Angular Speed = 0 + 3.3 rad/sec^2 * 16 sec = 52.8 rad/sec. This is the answer for part a).

For part b) you know the Initial Position = 0, Initial Angular Speed = 0, Angular Acceleration = 3.30 rad/sec, and Time = 16.0 sec. So solving the equation for part b) gives:
Final Angular Position = 0 + 0 + 1/2 * 3.3 rad/sec^2 * 16 sec^2 = 422.4 rads.

If you look at the equations I used in solving the problem you should notice they are very similar to the linear equations for speed and position. This should make them easier to remember.
Hope this helps.

Answer 2

V = A x t
V = 3.3 rad/s^2 (* 16s)
V = 52.8 rad/s
d = 1/2(Vf + Vi) x t
d = .5 * (52.8 + 0) x 16
d = 422.4 rad
pi (3.14) = 180 degrees
Angle = 422.4 rad x 180/3.14 rad = 24,214 degrees

Main thing to understand is that V = D / t ; D = (Vf + Vi) x t /2 [this says that if you should multiply your average velocity by time to get the distance, only if acceleration is constant] ; 2 pi = 360 deg ; and

Answer 3

Are you seriously using this to do your homework?