A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 48.0 revolutions before coming to rest. Find the constant angular acceleration (in rad/s2) of the centrifuge.

a = -210,066 rad/s2

Explanation:

The rotation speed of the centrifuge is 3400 revolutions/min. Converting this to rev/sec, we divide by 60, so 56.666 rev/sec

when switched off, it rotated 48 revolutions before stopping. using the Torricelli formula, we can find the acceleration it took to stop:

V2 = Vo2 + 2*a*DS, where:

V = 0

Vo = 56.666

DS = 48

Then, applying these values in the formula, we have:

0 = 3211.111 + 2*a*48

a = -3211.111/96 = -33.45 rev/s2

to convert from revolution to rad, we can multiply by 2*pi (1 revolution = 2pi rad)

a = -33.45 * 2 * pi = -210,066 rad/s2

Given Information:

Initial angular speed = ωi = 3,400 rev/min  = 56.67 rev/sec

Angular displacement = ΔΘ = 48 revolutions

Required Information:

Angular acceleration = α = ?

Angular acceleration = -209.41 rad/s²

Explanation:

We know from the kinematics,

2aS = Vf² - Vi²

This equation can be written in terms of rotational motion as

2αΔΘ = ωf² - ωi²

α = (ωf² - ωi²)/2ΔΘ

Where the final angular speed ωf² is zero since it was switch off.

α = (0 - (56.67)²)/2*48

α = (0 - (56.67)²)/2*48

α = -33.33 rev/s²

Since 1 revolution is equal to 2π radians

α = -33.33*2π

α = -209.41 rad/s²

The negative sign indicates deceleration.

Therefore, the constant angular acceleration of the centrifuge is 209.41 rad/s².

there are no contact forces mentioned on the list of choices that you've provided.