Question

A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a
centrifuge very carefully because of the very large accelerations.

(a) What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation?
(b) For comparison, what is the magnitude of the acceleration a test tube would experience
if dropped from a height of 1.0 m and stopped in a 1.0-ms-long encounter with a hard
floor?

Answer 1

Answer on Question #45360, Physics, Mechanics | Kinematics | Dynamics

A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations.

(a) What is the acceleration at the end of a test tube that is $10\mathrm{cm}$ from the axis of rotation?

(b) For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of $1.0\mathrm{m}$ and stopped in a 1.0-ms-long encounter with a hard floor?

Solution.

a)

The relation for the centripetal acceleration is:

$$a _ {C} = \omega^ {2} R = 4 \pi^ {2} f ^ {2} R$$

Where $f$ is a frequency of rotation, $R$ - distance from axis of rotation to the end of the tube.

Numerically:

$$a _ {C} = 4 \cdot 3. 1 4 ^ {2} \cdot \left(4 0 0 0 \frac {1}{6 0 s}\right) ^ {2} \cdot 0. 1 m \approx 1 7. 5 \cdot 1 0 ^ {3} \frac {m}{s ^ {2}}$$

b)

Just before hitting the floor tube has the speed:

$$V = g t$$

Where $t$ is the time of falling. It can be obtained from relation:

$$\frac {g t ^ {2}}{2} \rightarrow t \sqrt {\frac {2 h}{g}}$$ $$V \quad g \sqrt {\frac {2 h}{g}} \quad \sqrt {2 g h}$$

After hitting the floor speed of the tube is equal to 0, so acceleration is:

$$a \quad V / \Delta t$$

Where $\Delta t$ is the time of interaction of tube with the floor. So:

$$a \quad \frac {\sqrt {2 g h}}{\Delta t}$$

Numerically:

$$a \quad \frac {\sqrt {2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 1 m}}{0 . 0 0 1 s} \approx 4. 4 \cdot 1 0 ^ {3} \frac {m}{s ^ {2}}$$

Answer:

a) $a \approx 17.5 \cdot 10^{3} \frac{m}{s^{2}}$

b) $a \approx 4.4 \cdot 10^{3} \frac{m}{s^{2}}$

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