A wheel has angular acceleration of `3.0 rad//s^2` and an initial angular speed of `2.00 rad//s`. In a tine of `2 s` it has rotated through an angle (in radian) of

Options

A. 6
B. 10
C. 12
D. 4

Answer

B

Explanation

`omega = omega_(0) + alpha t …….(i)` becomes `(d theta)/(dt) = omega_(0) + alphat , ` i.e `d theta = (omega_(0)+alphat)dt`
So , if in time t angular displacement is `theta `
`int_(0)^(theta) d theta = int_(0)^(t) (omega_(0) + alphat) dt ` or
`theta = omega_(0)t +1/2t^(2)........(ii)`
Given ,
`alpha = 3.0 "rad/s"^(2) , omega_(0) = 2.0 "rad//s"t = 2s ` ,
Hence , `theta = 2 xx 12+ 1/2 xx 3xx (2)^(2) or theta = 4+6 = 10 ` rad