A centrifuge in a medical laboratory rotates at an angularspeed of 3600 rev/min.

Question

A centrifuge in a medical laboratory rotates at an angularspeed of 3600 rev/min. When switched off it rotates 50 times beforecoming to rest. What is the constant angular acceleration of thecentrifuge? A)-129600/s2 B)-113.1/s2 C)-226.2/s2 D)-36/s2 E)-5.73/s2 Hint: Formulas for rotational kinematics: =t    f=j+t f=i+i(tf-ti)+1/2(tf-ti)2 =<>t. <>=(i+f)/2 Given: i=0, f=50*2(50revolutions) i=(3600*2/min)*(1min/60s)=377/s,f=0 =constant Asked: =? Solution: Find t:<>=(i+f)/2=188.5/s, t=/<> Find : =(f-i)/t =<>t. <>=(i+f)/2 Given: i=0, f=50*2(50revolutions) i=(3600*2/min)*(1min/60s)=377/s,f=0 =constant Asked: =? Solution: Find t:<>=(i+f)/2=188.5/s, t=/<> Find : =(f-i)/t

Explanation / Answer

   Given Initial angular velocity i = 3600 rev/min = 376.8rad/s final angular velocity    f = 0rad/s and the angular displacement = 50 (2) rad = 314rad from rotational dynamics       f2 -i2 = 2 therefore angular acceleration = (f2 -i2) / 2 = - (376.8rad/s)2 /2(314 rad) = 226.08 rad/s2

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