An ultracentrifuge accelerates from rest to 100,000 rpm in 1.20 min.

Question

(a) What is its angular acceleration in rad/s^2?
(b) What is the tangential acceleration, in m/s^2, of a point 13.30 cm from the axis of rotation?
(c) What is the centripetal acceleration, in m/s^2, of this point at full rpm?
(d) Express this centripetal acceleration as a multiple of g.

Answer 1

Let's begin by converting the given values to appropriate units.

(a) To find the angular acceleration, we use the formula:

angular acceleration (α) = (final angular velocity - initial angular velocity) / time

Given that the final angular velocity is 100,000 rpm (revolutions per minute), we first convert it to radians per second by multiplying by 2π/60 (since there are 2π radians in one revolution and 60 seconds in a minute):

100,000 rpm * 2π/60 = 10,472 rad/s

The initial angular velocity is 0 as the ultracentrifuge starts from rest.

Plugging these values into the formula, we get:

α = (10,472 rad/s) / (1.20 min) = 8727 rad/s^2

Therefore, the angular acceleration of the ultracentrifuge is 8727 rad/s^2.

(b) To find the tangential acceleration at a point 13.30 cm from the axis of rotation, we use the formula:

tangential acceleration (at) = r * α

Given that r = 13.30 cm = 0.133 m, we have:

at = 0.133 m * 8727 rad/s^2 = 1170 m/s^2

Therefore, the tangential acceleration at that point is 1170 m/s^2.

(c) The centripetal acceleration at that point is given by the formula:

centripetal acceleration (ac) = r * ω^2

Substitute the values:

ac = 0.133 m * (10,472 rad/s)^2 = 1481 m/s^2

Therefore, the centripetal acceleration at that point at full rpm is 1481 m/s^2.

(d) To express this centripetal acceleration as a multiple of g, we divide by the acceleration due to gravity:

1481 m/s^2 / 9.81 m/s^2 = 151.03 times the acceleration due to gravity (g)

Therefore, the centripetal acceleration at that point is 151.03 times the acceleration due to gravity.