An ultracentrifuge accelerates from rest to 100,000 rpm in 2.50 min. (a) What is its angular acceleration in rad/s2? (b) What is the tangential acceleration (in m/s2) of a point 11.50 cm from the axis of rotation?
a) Angular acceleration:
First, convert the speed in rpm to rad/s as follows:
`omega` = 100,000 rev/min x 2`pi` rad/rev x 1 min/60 sec = 10,472 rad/sec
Since the ultracentrifuge starts from rest,
angular acceleration, `alpha` = `omega` /time
where time = 2.5 min = 2.5 min x 60 sec/min = 150 sec
Thus, angular acceleration, `alpha` = 10,472 rad/sec / 150 sec = 69.81 rad/`s^2`
b) Tangential acceleration can be calculated by using the following formula:
`a_t` = `alpha r`
where r is the distance from the axis of rotation. This distance is given as 11.50 cm.
In other words, r = 11.50 cm = 11.50 cm x 0.01 cm/m = 0.115 m.
Substituting the values of `alpha` and r in the above equation, we get:
`a_t` = (69.81 rad/s^2) x 0.115 = 8.03 m/`s^2`
Thus, the angular acceleration is 69.81 rad/`s^2`, and the tangential acceleration is 8.03 m/`s^2`.
Hope this helps.
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