Question

A wheel has a constant angular acceleration of 3.3 rad/s^(2). During a certain 6.0 s interval, it turns through an angle of 93 rad. Assuming that the wheel started from rest, how long had it been in motion before the start of the 6.0 s interval?

Answer 1

here we can write this ODE:
\[\frac{{d\omega }}{{dt}} = k\]
where \(k=3.3\)

Answer 2

let's suppose that the 6 seconds interval starts at time \(t_1\), then the angular speed and the angular displacement, at \(t_1\), are:
\[\omega \left( {{t_1}} \right) = k{t_1},\quad \theta \left( {{t_1}} \right) = k\frac{{t_1^2}}{2}\]
repsectively.
Now, at time \(t_1\) starts the 6 seconds interval, and final time is \(t_2\), namley:
\(t_2-t_1=6\)
so, integrating twice that ODE, with the initial condition above, we get:
\[\Delta \theta = \theta \left( {{t_2}} \right) - \theta \left( {{t_1}} \right) = k{t_2}\left( {\frac{{{t_2}}}{2} + {t_1}} \right)\]
where \( \Delta \theta =93\).
Remembering that \(t_2=t_1+6\), we can easily compute the value of \(t_1\)

Answer 3

oops.. namely*