Question

1. A turntable otates with constant a.as vad)s2 angul ar ace elevatonter 4.0D it has ota tud Harovgh n angle f GD.Dvd What wa the angular velb oty f he wheel at the begnning of the 4.oD Snterval 7 ,寺 4.00S

Answer 1

let us assume angular velocity at beginning of 4-sec interval be Wu.
There let q = 60 rad be angle rotated during this 4-sec interval.


Since angular acceleration of table j = 2.25 r/s^2 is constant then expression for q can be found as-
q = Wu*t + (1/2)*j*t^2 ; this is similar to linear velocity
So s = U*t + (1/2)*a*t^2

putting the values we get
60 = Wu*4 + (1/2) * 2.25 * 4^2
or Wu = (60 - 18) / 4
Wu = 10.5 rad/s