The figure below shows an arrangement of 5 identical disks that have been glued

Question

The figure below shows an arrangement of 5 identical disks that have been glued together in a rod-like shape of length L = 1.35 m and (total) mass M = 90 grams. The arrangement is rotated through its central disk at point O (around an axis perpendicular to the page).

The figure below shows an arrangement of 5 identical disks that have been glued together in a rod-like shape of length L = 1.35 m and (total) mass M = 90 grams. The arrangement is rotated through its central disk at point O (around an axis perpendicular to the page). What is the rotational inertia of the arrangement about that axis? kg middot m2 If we approximated the arrangement as being a uniform, very thin rod of mass M and length L, how far would we be off from the exact rotational inertia? Express as a percent error. You may answer as a magnitude (+).

Explanation / Answer

rotational inertia of the arrangement about that axis

0.5*mR^2 + 2*(0.5*mR^2 + m(2R)^2) +2*(0.5*mR^2 + m(4R)^2) 42.5mR^2 m

= M/5 R=L/10 =0.425ML^2

Of Rod of mass M and Length L

= 0.33ML^2

Percentage Error

= {0.425-0.33)/0.33}*100 28.78%

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