Question

In: Physics

1. A centrifuge rotor is accelerated for 28. s from 9000 rpm to 17000 rpm (revolutions...

1. A centrifuge rotor is accelerated for 28. s from 9000 rpm to 17000 rpm (revolutions per minute).

(a) What is its average angular acceleration?

2. A centrifuge rotor is accelerated for 20. s from 7000 rpm to 15000 rpm (revolutions per minute).

(a) What is its average angular acceleration?

Expert Solution

1.

Using 1st rotational kinematic equation:

wf = wi + *t

= (wf - wi)/t

wi = Initial angular velocity = 9000 rpm = 9000*2*pi/60 = 942.48 rad/sec

wf = final angular velocity = 17000 rpm = 17000*2*pi/60 = 1780.24 rad/sec

t = 28 sec

So,

= (1780.24 - 942.48)/28

= 29.92 rad/sec^2 = 30.0 rad/sec^2 = average angular acceleration

1.

Using 1st rotational kinematic equation:

wf = wi + *t

= (wf - wi)/t

wi = Initial angular velocity = 7000 rpm = 7000*2*pi/60 = 733.04 rad/sec

wf = final angular velocity = 15000 rpm = 15000*2*pi/60 = 1570.8 rad/sec

t = 20 sec

So,

= (1570.8 - 733.04)/20

= 41.888 rad/sec^2 = 41.9 rad/sec^2 = average angular acceleration

Let me know if you've any query.

genius_generous answered 2 years ago

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