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Author Question: A uniform rod of mass M = 1.2 kg and length L = 0.80 m, lying on a frictionless horizontal plane, is ... (Read 239 times)

amal

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on: Aug 1, 2020
A uniform rod of mass M = 1.2 kg and length L = 0.80 m, lying on a frictionless horizontal plane, is free to pivot about a vertical axis through one end, as shown. The moment of inertia of the rod about this axis is given by (1/3)ML2. If a force (F = 5.0 N, θ = 40°) acts as shown, what is the resulting angular acceleration about the pivot point?



◦
16 rad/s2

◦
12 rad/s2

◦
14 rad/s2

◦
10 rad/s2

◦
33 rad/s2
Physics for Scientists and Engineers ¦ Serway, Jewett ¦ 9th Edition
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Marked as best answer by amal on Aug 1, 2020

meganmoser117

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Reply #1 on: Aug 1, 2020
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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amal

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Reply 2 on: Aug 1, 2020
:D TYSM

ASDFGJLO

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Reply 3 on: Yesterday
Great answer, keep it coming :)

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