rotational kinematics?

Question

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplished by spinning a sample around in a circle with a large angular speed. Suppose that after a centrifuge in a medical laboratory is turned off it continues to rotate with a constant angular deceleration for 24.7 s before coming to rest.
(a) If its initial angular speed was 2870 rpm, what is the magnitude of its angular deceleration?
...rev/s2
(b) How many revolutions did the centrifuge complete after being turned off?
...rev

Answer 1

The equations you need are

ω = ωo + αt
θ - θo = ωo t + 1/2 α t²

where θ is the final position in radians ( unknown ), θo is the initial position ( zero ), ω is the final angular velocity ( zero ), ωo is the initial angular velocity ( 2870 rpm converted to rad / sec ), α the angular acceleration ( unknown ), and t the elapsed time ( 24.7 sec ).

2870 rpm = ( 2π ) ( 2870 ) / ( 60 ) radians / second
Use the first equation to solve for the angular acceleration α ( it will be negative ), then use that value in the second expression to find the final position θ. Divide that by 2π to convert the answer from radians to revolutions.

Answer 2

For the 1st situation, the 2nd hand will return and forth a million radian for each 6 cm of arc. mentioned yet in any different case, for an attitude of a million radian, the arc length equals the radius. A radian is approximately 57 levels. 24.0 / 6.00 = 4 radians all of us understand 360 levels = 2 pi radians 4 radians * (360 deg / 2 pi radians) * (60 seconds / 360 levels) = 38.2 seconds For the 2nd situation, i'm utilising Latin "equivalents" to the Greek letters omega and alpha ?- angular speed a- angular acceleration ? = ?(0) + ?t ? = 0 + ?t sixteen = ?7 ? = sixteen / 7 14 = (sixteen/7) t t = 6.a million seconds For the 0.33 situation: ?- no longer relatively defined, angular displace is a subtraction of one ? from yet another ?- angular speed ?- angular acceleration ? = ?(0) + ?(0)t + ½?t² because of the fact it starts at relax, the preliminary angular speed is 0. We mentioned until eventually now that angular displace is the subtraction of two ?. we can adjust the above equation to ? - ?(0) = ?(0)t + ½?t² and then say displacement is ? - ?(0). We eradicated the preliminary speed term (ranging from relax), leaving us with: ? - ?(0) = ½?t² 22 = ½ ? 2.5² 7.04 = ? ? - ?(0) = ½(7.04)5² ? - ?(0) = 88 rad

Answer 3

rotational kinematics are just like linear kinematics. The same equations apply, although the variables are different.

so

v= vo + a*t linear
omega=omega o + alpha*t rotational

for the second part, use equivalent equation for postion in linear kinematics for rotational kinematics.