The angular acceleration a of a spinning top as a function of t is : `alpha = 3t^(2) + 5t`. At t = 0, the angular velocity `omega_(0) = 10` rad/s and angular position `theta = 8` rad. The angular position as a function of time t is given by which of the following expression?

Options

A. `(1)/(4) t^(4) + (5)/(6) t^(3) + 10 t + 8`
B. `(5)/(6) t^(4) + (1)/(4) t^(3) + (2)/(5) t + 8`
C. `2t^(4) + 3t^(3) + 5t + 8`
D. `(1)/(4) t^(4) + (3)/(4) t^(3) + 6t^(2) + 8`

Answer

A

Explanation

Given : `alpha = 3t^(2) + 5t`
`alpha = (d^(2) theta)/(dt^(2)) = 3t^(2) + 5t " " … (i)`
`alpha = (d^(2) theta)/(dt^(2)) = 3t^(2) +5t " " …. (i)`
Integrate both sides w.r.t. t in equation (i) , we get
`(d theta)/(dt) = t^(3) + (5)/(2) t^(2) + C_(1)`
where `C_(1)` is a constant of integration .
or `omega = (d theta)/(dt) = t^(3) + (5)/(2) t^(2) + C_(1)`
Using initial conditions , at t = 0 , `omega_(0) = 10` rad/s
`therefore C_(1) = 10` rad/s
or `(d theta)/(dt) = t^(3) + (5)/(2) t^(2) + 10 " " .... (ii)`
Integrate both sides w.r.t t in equation (ii) , we get
`theta = (1)/(4) t^(4) + (5)/(6) t^(3) + 10 t + C_(2)`
where `C_(2)` is a constant of integration .
Using initial conditions , at t = 0 , `theta_(0) = 8` rad
`therefore C_(2) = 8` rad or `theta = (1)/(4) t^(4) + (5)/(6) t^(3) + 10 t + 8`