The shaft of a pump starts from rest and has an angular acceleration of 2.00 rad/s2 for 15.0 s.
(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?
2006-11-13 03:39:29 · 2 answers · asked by Alan l 1 in Science & Mathematics ➔ Physics
(a) too trivial to answer
(b) Take the average speed times the number of seconds over which it had that average speed.
2006-11-13 03:54:14 · answer #1 · answered by actuator 5 · 0⤊ 0⤋
a) w = aa t; where w = angular velocity (rad/sec), aa = angular acceleration (2 rad/sec^2), and t = time interval of acceleration (15 sec).
b) W = avg(w) t; where avg(w) = w/t; where W = the angle (rad) the shaft moved at avg(w) average angular velocity over the period (t).
In terms of a circle a radian (rad) can be seen as the ratio of the length of the arc subtended by two radii to the radius of the circle. If L = C = 2 pi R; where L is the length of arc equal to the circumference (C) of the circle, then rad = C/R = 2 pi R/R by definition; so that rad = 2 pi which is equivalent to 360 degrees.
With the equations and rad definition, you can do the math.
2006-11-13 04:19:43 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋