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?

Expert Answers

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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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