Question

A centrifuge consists of a container held at a distance of 0.20 m from its axis. When turned on, the centrifuge spins the container and its contents round at 9000 revolutions per minute. a) Find its angular velocity. b) A small mass of 10 g is placed in the drum. Work out what force will need to be provided for it to go round at that rate in a circle of radius 0.20 m. c) In which direction must that force act? d) what supplies that force to the ball? e) taking g to be 9.8N/kg, what size force will the earth supply to the ball when it is at rest on the floor?​

Answer 1

Hi there!

a)

We can convert 9000 rev/min to rad/sec (proper units for angular velocity) using dimensional analysis.

$\frac{9000rev}{min} * \frac{min}{60 s}* \frac{2\pi rad}{1 rev} = \boxed{942.48 rad/sec}$

b)

For the mass to go around at this speed, we can use the rotational equivalent for centripetal force:

$F_c = m\omega ^2r$

Plug in the given values. Remember to convert grams to kg.

$F_c = (0.01)(942.48^2)(0.20) = \boxed{1776.53 N}$

c)

The force is a CENTRIPETAL force, so it must act towards the CENTER of the circle at all times.

d)

The only force working in the horizontal direction on the mass is:

$\Sigma F = F_c = F_N$

The normal force from the container's walls provide the centripetal force experienced by the mass.

e)

When the ball is at rest, the force supplied by the earth is equivalent to its weight:

$W = mg$

Plug in the givens:

$W = (0.01)(9.8) = \boxed{0.098 N}$