1.) A centrifuge in a medical laboratory rotates at an angular speed of 3,500 re

Question

1.) A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min. When switched off, it rotates through 52.0 revolutions before coming to rest. Find the constant angular acceleration (in rad/s2) of the centrifuge.

2.)The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 7.0 rev/s in 8.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 20-s interval? Assume constant angular acceleration while it is starting and stopping.

(a) How long does it take for the grinding wheel to stop?
s

(b) Through how many radians has the wheel turned during the interval found in (a)?
rad

4.) A sample of blood is placed in a centrifuge of radius 17.0 cm. The mass of a red blood cell is 3.0 1016 kg, and the magnitude of the force acting on it as it settles out of the plasma is 4.0 1011 N. At how many revolutions per second should the centrifuge be operated?

Explanation / Answer

According to the given problem,

1.) Employ the following rotational kinematic equation:

2 = 02 - 2

where, final angular velocity: = 0,initial angular velocity: 0 = (2/60)n = (2/60)(3500) = 366.52 rad/s, angular acceleration: angular displacement: =52(2) = 326.725 rad

2 = 0 = 02 - 2,

= 02 / 2 = 366.522 / 2*301.6

= 205.58 rad/s2

2.) Assuming constant angular acceleration while it is starting and stopping.

SPEED UP
Velocity 0 = 0 rev/s
Velocity = 7 rev/s
Time t = 8s

= 1/2*( 0 + )*t
= 1/2*( 0 + 7 )*8
= 28 rev

SLOW DOWN
Velocity 0 = 7 rev/s
Velocity = 0 rev/s
Time t = 12s

= 1/2*(0+ )*t
= 1/2*(7 + 0)*12
= 42 rev

Total revolutions in 20 seconds:
28 + 42 = 70 rev

3.)For Constant Deceleration,

(a) Using the rotaional kinamatic equations,

= 0 + t

where t = time to stop, = 0 = - 1.70 rad/s2, 0 = 106*[2/60] = 11.1003 rad/s

= 0 + t
0 = 11.1003 rad/s - 1.7rad/s2*t
t = 11.1003/1.7 = 6.53sec

(b) Using the rotaional kinamatic equations

= 1/2*(0+)*t where,0 = 11.003 rad/sec, = 0, t = 6.53 sec
= 1/2 (0+11.003 rad/sec) * 6.05 sec
= 35.922 rad

4.)According to the given problem,

Centripetal force F = mV2/R, rearrage to get velocity

V= FR/m = (4*10-11*0.17)/3*10-16

V = 150.554 m/s

The distance travlled is circumference = 2R = 2*0.17 = 1.068 m

So single revolution takes 1.068 / 163.3 = 0.00709s

or 1/0.00769 = 140.94949 rev/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.