What will be the angular acceleration of the turntable plus the ring a_iR or the

Question

What will be the angular acceleration of the turntable plus the ring a_iR or the disk a_td when the experiment is run again? To work this out. use Eq. (7) for the angular acceleration a and substitute l_T + l_D (or I, + l_R) for/. Be careful about units and use the correct values of m and r_v Now carry out measurements and calculations to test the prediction. Follow the steps used previously to measure the angular acceleration, find a mean and standard deviation of the mean of a. Put your results in the tables below.

Explanation / Answer

mean is

u = 7.6746+7.6522+7.7225+7.6896+7.8305/5 =7.71388

( xi-u)^2

(7.6746-7.71388)^2 = 0.0015429184

(7.6522-7.71388)^2 = 0.0038044224

(7.7225-7.71388)^2 =0.0000743044

(7.6896-7.71388)^2 = 0.0005895184

(7.8305-7.71388)^2 = 0.0136002244

standard devitation 1/N ( sum of ( xi- u)^2

standard devitation 1/5 ( 0.019611388) = 0.0039222776

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mean = 10.315 + 10.329+10.332+10.326 +10.400/5 = 10.3384

( 10.315-10.3384)^2 = 0.00054756

(10.329-10.3384)^2 =0.00008836

(10.332-10.3384)^2 =0.00004096

(10.326-10.3384)^2 =0.00015376

(10.4-10.3384)^2 =0.00379456

standard devitation

1/N ( sum of xi-u)^2

1/5 ( 0.0046252 ) = 0.00092504

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