Answer to Question #99992 in Classical Mechanics for William Huddleston

Question #99992
A steel rod 1 M long and having a mass of 2 kg is pinned at one of its ends. A force is applied perpendicularly to the rod at a distance of 75 cm from the pinned end. What is the magnitude of the resulting angular acceleration of the rod? Moment of inertia is 2 kg m^2
1
Expert's answer
2019-12-06T10:06:49-0500
"F=W=mg"

Total moment is


"M=-W(0.5L)+Fx=-0.5mgL+mgx"

"M=(2)(10)(0.75-0.5(1))=5\\ Nm"

The resulting angular acceleration of the rod:


"\\alpha=\\frac{M}{I}=\\frac{5}{2}=2.5\\frac{rad}{s^2}"


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