Question

A centrifuge rotor rotating at 9000 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.73 m⋅N

Part A: If the mass of the rotor is 3.72 kg and it can be approximated as a solid cylinder of radius 0.0760 m, through how many revolutions will the rotor turn before coming to rest?
Express your answer to three significant figures.

Part B: How long will it take?
Express your answer to three significant figures and include the appropriate units.

Answer 1

Look below

Step-by-step explanation:

Part A:

The initial angular velocity of the rotor can be found using the formula:

ω = v/r

where ω is the angular velocity, v is the linear velocity, and r is the radius of the cylinder. Since the rotor is rotating at 9000 rpm, we can convert this to radians per second:

ωi = (9000 rpm) x (2π/60 s) = 942.48 rad/s

The final angular velocity is zero. The frictional torque acting on the rotor is given by:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The moment of inertia for a solid cylinder is:

I = (1/2)mr^2

Substituting in the given values and solving for α, we get:

α = τ/I = (1.73 m⋅N) / [(1/2)(3.72 kg)(0.0760 m)^2] = 371.7 rad/s^2

The final angular velocity is zero, so we can use the formula:

ωf^2 = ωi^2 + 2αθ

where θ is the angle of rotation. Solving for θ, we get:

θ = (ωf^2 - ωi^2) / (2α) = (0 - (942.48 rad/s)^2) / (2 x 371.7 rad/s^2) = 6.76 revolutions

Therefore, the rotor will turn approximately 6.76 revolutions before coming to rest.

Part B:

The time it takes for the rotor to come to rest can be found using the formula:

ωf = ωi + αt

where ωf is the final angular velocity, ωi is the initial angular velocity, and α is the angular acceleration. Solving for t, we get:

t = (ωf - ωi) / α = (0 - 942.48 rad/s) / 371.7 rad/s^2 = 2.535 s

Therefore, it will take approximately 2.535 seconds for the rotor to come to rest.