Hi there. Can you guys please solve both problems shown in the pictures within 2

Question

Hi there. Can you guys please solve both problems shown in the pictures within 2 hours? Please explain and show all the steps in details for a clear understanding.
Thank you very much!

”40 Figure 10-33 shows an arrangement of 15 identical disks that have been glued together in a rod-like shape of length L 1.0000 m and (total) mass M- 100.0 mg. The disk arrangement can rotate about a perpendicular axis through its central disk at point O. (a) What is the rotational inertia of the arrangement about that axis? (b) If we approximated the arrangement as being a uniform rod of mass M and length L, what percentage error would we make in us- ing the formula in Table 10-2e to calculate the rotational inertia? Fig. 10-33 Problem 40.

Explanation / Answer

40)Given,

n = 15 ; L = 1 m ; M = 100 mg ;

a)mass of each disk

m = 100/15 = 6.67 mg

radiys of each disk

r = 1/15/2 = 0.033 m

moment of inetrtia od a single disk

I = 1/2 m r^2

I = 0.5 x 6.67 x 10^-3 x 0.033^2 = 3.6 x 10^-6 kg-m^2

Using parallel axis theorem

I = 15 x 3.6 x 10^-6 + 6.67 x 10^-3 x 0.033^2 (2 x 2^2 + 2 x 4^2 + 2 x 6^2 + 2 x 8^2 + 2 x 10^2 + 2 x 10^2 + 2 x 14^2) = 0.0076 kg-m^2

Hence, I = 0.0076 kg-m^2

b)I(rod) = 1/12 M L^2

I(rod) = 100 x 10^-3 x 1^2/12 = 0.0083

Hence, I = 0.0083 kg-m^2

% error = (0.0083 - 0.0076)/0.0083 x 100 = 8.43%

Hence, %age error = 8.43 %

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