High School

An ultracentrifuge accelerates from rest to 10, 500 rad/s in 2.00 min. how many radians did it rotate for the 2.00 min. time interval?

Answer :

The ultracentrifuge rotated approximately 630,000 radians during the 2.00min time interval. This can be calculated using the formula θ = ωi × t + (1÷2) × α × [tex]t^{2}[/tex], where ωi is the initial angular velocity (0 rad[tex]s^{-1}[/tex]), t is the time interval (2.00 min or 120s ), and α is the angular acceleration.

θ = ωi × t (1÷2) × α × [tex]t^{2}[/tex],

in which θ is the angular displacement, ωi is the preliminary angular velocity, t is the time c program language period, and α is the angular acceleration.

Given:

ωi = 0 rad[tex]s^{-1}[/tex] (initial angular pace)

ωf = 10,500 rad[tex]s^{-1}[/tex] (final angular pace)

t = 2.00 min = 120s (time c program language period)

We need to locate θ.

First, allow calculate the angular acceleration (α) the use of the components:

α = (ωf - ωi) ÷ t.

α = (10,500 rad[tex]s^{-1}[/tex] - 0 rad[tex]s^{-1}[/tex]) ÷ 120 s = 87.5 rad[tex]s^{-2}[/tex]

Now, we can calculate the angular displacement (θ) the use of the method:

θ = ωi × t (1÷2) × α [tex]t^{2}[/tex]

θ = 0 rad[tex]s^{-1}[/tex] × 120s (1÷2) × 87.5 rad[tex]s^{-2}[/tex] × (120)[tex]s^{2}[/tex].

θ = 0 (1/2) × 87.5 rad[tex]s^{-2}[/tex] * 14, 400[tex]s^{2}[/tex],

θ = (1/2) × 87. 5rad[tex]s^{-2}[/tex] × 14,400[tex]s^{2}[/tex],

θ = 43.75 rad[tex]s^{-2}[/tex] × 14,400[tex]s^{2}[/tex].

θ ≈ 630,000 radians.

Therefore, the ultracentrifuge rotated about 630,000 radians during the two.00-minute time c language.

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