Question

An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00 min. (a) what is its angular acceleration in rad/s2 ? (b) what is the tangential acceleration of a point 9.50 cm from the axis of rotation? (c) what is the radial acceleration in m/s2 and multiples of ???? of this point at full rpm?

Answer 1

(a) The angular acceleration is given by:

$\alpha=\frac{\omega_f-\omega_i}{\Delta t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

$\Delta t$ is the time interval

Let's convert the angular velocities in rad/s and the time in seconds:

$\omega_f=100000 \frac{rev}{min}=100000 \frac{rev}{min} \frac{2 \pi rad/rev}{60 s/min}=10467 rad/s$

$\omega_i=0$

$\Delta t=2.00 min \cdot 60 \frac{s}{min} = 120 s$

Substituting the numbers into the equation, we find the angular acceleration:

$\alpha=\frac{10467 rad/s}{120 s}=87.2 rad/s^2$

(b) The tangential acceleration is given by the product between the angular acceleration and the distance of the point from the axis of rotation. In this case, the distance is

$r=9.5 cm=0.095 m$

Therefore the tangential acceleration is

$a=\alpha r=(87.2 rad/s^2)(0.095 m)=8.3 m/s^2$

(c) The radial acceleration (also known as centripetal acceleration) is given by the product between the square of the angular velocity and the distance from the axis of rotation:

$a=\omega^2 r=(10467 rad/s)^2(0.095 m)=1.04 \cdot 10^7 m/s^2$