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A centrifuge in a medical laboratory rotates at a rotational speed of 3600 rev/min,...

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kjfuefsap | eNotes Newbie

Posted November 12, 2013 at 6:22 PM via web

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A centrifuge in a medical laboratory rotates at a rotational speed of 3600 rev/min, When switched off, it rotates 50.0 times at a constant angular acceleration before coming to rest.

a) Determine the initial angular speed of the centrifuge.

b) Determine the angle (in radians) through which the centrifuge rotates  before coming to rest.

c) Calculate the constant angular acceleration of the centrifuge.

d) Calculate the time necessary for the centrifuge to come to rest.

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted November 12, 2013 at 7:51 PM (Answer #2)

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a) The initial angular speed is simply

`omega_0 = 2*pi*F =2*pi*3600 rad/min =2*pi*3600/60 (rad)/(sec)=`

` =120*pi (rad)/(sec) ~~377 (rad)/(sec)`

The initial angular speed is `120*pi` rad/sec

b)

The centrifuge does 50 complete rotations before it stops. One complete rotation means an angle of `2*pi` . The total angle of rotation is thus

`alpha =50*(2*pi) =100*pi =314.16 rad `

or `alpha =50*360 = 18000 degree`

The total angle until stop is `100*pi` rad.

c) The angular acceleration comes from

`omega^2 =omega_0^2 +2*epsilon*alpha`

(similar to `V^2 =V_0^2+2*a*s` )

`omega =0`and `omega_0 =377 (rad)/s`  means

`epsilon = (omega^2-omega_0^2)/(2*alpha) =-377^2/(2*314.16) =-226.2 (rad)/s^2`

The angular acceleration is `-226.2 (rad)/sec^2`

d)

The time until stop comes from equation

`omega =omega_0 +epsilon*t`  (similar `V=V_0 +a*t` )

`t = -omega_0/epsilon =377/226.2 =1.67 sec`

The time until stop is 1.67 sec

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