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3 years ago

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2. A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point

located 8.00 cm from the axis of rotation of the centrifuge moves with a speed of 150 m/s when the centrifuge is at full speed. (a) What is the angular acceleration (in rad/s2) of the centrifuge as it spins up

2 answers:

irina1246 [14]3 years ago

4 0

Answer:

18.75 rad/s²

Explanation:

Given that

Mass of the centrifuge, m = 3.45 kg

Time taken to spin, t = 100 s

Distance from axis of rotation, r = 8 cm = 0.08 m

Speed of the centrifuge, v = 150 m/s

First, we find the angular velocity

Angular velocity, w = v / r

w = 150 / 0.08

w = 1875 rad/s

And from the angular velocity, we get our angular acceleration.

Angular acceleration = angular velocity / time taken

Angular acceleration = 1875 rad/s / 100 s

Angular acceleration = 18.75 rad/s²

Therefore, the angular acceleration is 18.75 rad/s²

Yuki888 [10]3 years ago

3 0

Answer:

The angular acceleration of the centrifuge as it spins up is 18.75 rad/s²

Explanation:

Given;

mass of centrifuge, m = 3.45 kg

time taken to spin, t = 100 s

distance from the axis of rotation, r = 8.00 cm = 0.08 m

final velocity of the centrifuge, v = 150 m/s

initial velocity of the centrifuge, u = 0

Determine the linear acceleration of the centrifuge at the given time;

$a = \frac{v-u}{t} \\\\a = \frac{150-0}{100} \\\\a = 1.5 \ m/s^2$

Finally, determine the angular acceleration of the centrifuge as it spins up;

α = a/r

where;

α is the angular acceleration

α = 1.5 / 0.08

α = 18.75 rad/s²

Therefore, the angular acceleration (in rad/s2) of the centrifuge as it spins up is 18.75 rad/s²

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