A torque of 1.45 N-m is applied to a thin rod, fixed to rotate about a point through its center. If the rod is 2.50 kg in mass and 1.11 m in length, find the resulting angular acceleration (in rad/s2) of the rod. (The moment of inertia for a thin rod rotating about a point through its center is /= ML2/12)
A torque of 1.45 N-m is applied to a thin rod, fixed to rotate about a point through its center. If the rod is 2.50 kg in mass and 1.11 m in length, find the resulting angular acceleration (in rad/s2) of the rod. (The moment of inertia for a thin rod rotating about a point through its center is /= ML2/12)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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![A torque of 1.45 N-m is applied to a thin rod, fixed to rotate about a point through its center. If the rod is 2.50 kg in mass and 1.11 m in length, find the resulting angular acceleration (in
rad/s2) of the rod. (The moment of inertia for a thin rod rotating about a point through its center is I = ML2/12)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8349a40f-6ce2-46de-8796-b63c75b713ac%2F81571cd7-c3ee-48e9-bf6e-5e0cd43571c9%2F2ih59wh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A torque of 1.45 N-m is applied to a thin rod, fixed to rotate about a point through its center. If the rod is 2.50 kg in mass and 1.11 m in length, find the resulting angular acceleration (in
rad/s2) of the rod. (The moment of inertia for a thin rod rotating about a point through its center is I = ML2/12)
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