Question

The tip of a spinning flagella moves in a circular path and has a initial angular velocity of +1.38 Hz and an angular acceleration of +2.91 s^-2. After 2.3 s of motion, determine the magnitude of the displacement (not the tangential displacement) of the tip of the flagella if the radius of it's path is 34 nm.

For simplicity, you may wish assume that the tip of the flagella initially begins on the x-axis.

Report your answer in nm.

Answer 1

Solution :

Given :

angular velocity () = 1.38 Hz

angular acceleration () = 2.91 s^-2

time interval (t) = 2.3 s

radius of path (r) = 34 nm

Since, angular displacement: = t + (1/2) t²

Thus : = (1.38 Hz)(2.3 s) + (0.5)(2.91 s^-2)(2.3 s)² = 10.87 rad

Hence, Number of revolutions made = 10.87 rad / (2) = 1.73 revolutions

Here, displacement for a complete revolution will be zero.

Therefore, Net displacement = (0.73 revolution) x (34 nm) = 58.82 nm