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An ultracentrifuge is spinning at a speed of 80,000 rpm. The rotor that spins with
the sample can be roughly approximated as a uniform cylinder of 10 cm radius
and 8 kg mass, spinning about its symmetry axis). In order to stop the rotor in
under 30 s from when the motor is turned off, find the minimum braking torque
that must be applied.
O-19.2 Nm
-17.2 Nm
O -15.2 Nm
O-11.2 Nm
O None of the above

An Ultracentrifuge Is Spinning At A Speed Of 80000 Rpm The Rotor That Spins With The Sample Can Be Roughly Approximated As A Uniform Cylinder Of 10 Cm Radius An class=

Sagot :

onyebuchinnaji

D. The minimum braking torque that must be applied is -11.2 Nm.

Moment of inertia of the uniform cylinder

The moment of inertia of the uniform cylinder is calculated as follows;

I = ¹/₂MR²

where;

  • M is mass of the cylinder
  • R is radius of the cylinder

I = (0.5)(8)(0.1²)

I = 0.04 kgm²

Minimum braking torque

τ = -Iα

where;

  • α is angular acceleration

α = ω/t

α = (80,000 x 2π/rev x 1 min/60s) / (30 s)

α =  (80,000 x 2π)/(60 x 30)

α = 279.25 rad/s²

τ = - ( 0.04 kgm²) x (279.25)

τ =  -11.2 Nm

Thus, the minimum braking torque that must be applied is -11.2 Nm.

Learn more about minimum braking torque here: https://brainly.com/question/6845979

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