A centrifuge in a medical laboratory rotates at an
angularspeed of 3800 rev/min. When switched off, it rotates through
46.0revolutions before coming to rest. Find the constant
angularacceleration of the centrifuge.

Answer :

Answer:

- 273.77 rad/s^2

Explanation:

fo = 3800 rev/min = 3800 / 60 rps = 63.33 rps

f = 0

ωo = 2 π fo = 2 x 3.14 x 63.33 = 397.71 rad/s

ω = 2 π f = 0

θ = 46 revolutions = 46 x 2π radian = 288.88 radian

Let α be the angular acceleration of the centrifuge

Use third equation of motion for rotational motion

[tex]\omega^{2}=\omega _{0}^{2}+2\alpha \theta[/tex]

[tex]0^{2}=397.71^{2}+2 \times \alpha \times 288.88[/tex]

α = - 273.77 rad/s^2

Other Questions