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12.11.3 Activity: The Rotational Inertia of a Disk Measure the radius, R, of the disk shown in Figure 12.17. Use the theoretical equation obtained rom integration to calculate the theoretical value of the rotational inertia of that disk. The a. theoretical equation isD2b You will need to find an equation for the area of a hoop or radius, r, in order to determine what fraction of the total mass of the disk is contained in each hoop. If the area of a disk of radius, r, is given by A r, show that the area of the first inner disk, is given by Al-Th: Since there are 10 hoops of equal "width",r,/10 b. 0,026 l O 2.1 xlo-S c. Show that the area of the second hoop can be calculated by subtracting the area of the inner disk from the area of the second disk so that A,-r, where r,-2(/10). lae ,(フィ102e さ乙 2 (o d. Show that the area of the n't hoop is given by subtracting the area of the (n 1)st disk from the nth disk so that Δ4n-tr2,-".1 where r"-n(r/10), and r-,-(n-1)(r/10). 1 6 lo UNIT 12 ROTATIONAL MOTION Create a spreadsheet with the values of n,r, and A4, for each hoop along with the value of rotational inertia contributed by each of the hoops. Affix the spreadsheet in the space following. Hint: If the disk has a uniform density, then the mass of each hoop, m, is proportional to its area, Δ4. so that m..(M.14,ju,where A,-tr' e. How closely does your numerical spreadsheet calculation for the total rotational inertia of the disk compare with the value you calculated theoretically? Write down the general equation for the percent discrepancy and display the steps in your calculation of it below t. How could you change your procedures to make the percent discrepancy smaller? z. i) 0 1999 John Wilcy & Sons Portions of this material may have bom modifiod locally Revised March 5, 2019

Answer 1

a. given
   R = 0.026 m
   for mass M
   theoretical moment of inertia is given by
   I = 0.5MR^2 = 3.38M*10^-4 kg m^2

b. given
   r1 = r/10 [thickness of each hoop]
   then the innermost part is again a disc
   hence
   area of innermost part is given by
   A1 = pi(r1)^2

c. area of second hoop shall be calculated as
   A2 = pi(r2)^2 - pi(r1)^2
   but r2 = 2*r1
   A2 = 3*pi*r1^2

d. hence from the previous part we can say
   An = pi(rn)^2 - pi(r(n-1))^2
   now
   rn = n(r/10)
   r(n-1) = (n-1)(r/10)
   hence
   An = pi(n^2*r^2/100 - (n-1)^2*r^2/100)
   An = pi(n^2 - (n-1)^2)*r^2/100
   An = pi(2n - 1)*r^2/100

e. for uniform density disc
   An = pi(2n - 1)*r^2/100
   r = 0.026 m
   also
   mass Mn = M*pi*(2n - 1)r^2/100*pi*r^2
   mass Mn = M*(2n - 1)/100

   let M = 1 kg
   then moment of inertia In = Mn*Rn^2 = (2n - 1)*n^2*r^2/100,00
   then from the following spreadsheet

   n In
   1   6.76E-08
   2   8.112E-07
   3   0.000003042
   4   7.5712E-06
   5   0.00001521
   6   2.67696E-05
   7   4.30612E-05
   8   0.000064896
   9   9.30852E-05
   10   0.00012844
  
   total moment of inertia is obtained by adding all the individual ones
   It = 0.000382954

f. from the given data
   for M = 1 kg
   theoretical moment of inertia is I = 3.38*10^-4
   hence percent error is (3.38*10^-4 - 0.000382954)*100/3.38*10^-4 = 13.3% error

g. the percent accuracy can be improved by taking more number of rings (hoops) while summing up the hoops to get final moment of inertia of the disc