Question

A centrifuge rotor rotating at 9800 rpmrpm is shut off and is eventually brought uniformly to rest by a frictional torque of 2.22 m⋅Nm⋅N .

Part A

Please Express your answer to three significant figures.

If the mass of the rotor is 3.67 kgkg and it can be approximated as a solid cylinder of radius 0.0640 mm , through how many revolutions will the rotor turn before coming to rest?

Part B

Express your answer to three significant figures and include the appropriate units.

How long will it take?

Answer 1

Initial rotational speed of the centrifuge = N₁ = 9800 rpm

Initial angular speed of the centrifuge = ₁

₁ = 1026.25 rad/s

Final angular speed of the centrifuge = ₂ = 0 rad/s (Comes to rest)

Mass of the rotor = M = 3.67 kg

Radius of the rotor = R = 0.064 m

Moment of inertia of the rotor = I

I = MR²/2

I = 7.516 x 10^-3 kg.m²

Frictional torque on the rotor = T_f = -2.22 N.m (Negative as it opposes the motion of the rotor)

Angular acceleration of the rotor =

I = T_f

(7.516x10^-3) = -2.22

= -295.37 rad/s²

Angle through which the centrifuge rotor rotates before coming to rest =

₂² = ₁² + 2

(0)² = (1026.25)² + 2(-295.37)

= 1782.83 rad

Number of revolutions made by the rotor before coming to rest = n

n = 284 rev

Time taken for the rotor to come to rest = T

₂ = ₁ + T

0 = 1026.25 + (-295.37)T

T = 3.47 sec

A) Number of revolutions the rotor will turn before coming to rest = 284 rev

B) Time taken for the rotor to come to rest = 3.47 sec