A thin uniform rod 2.40 m long weighing 135 N has a frictionless pivot at one end, allowing it to rotate in a vertical plane. If it is rotated so that it is vertical (with the pivot at the bottom) and released, the force of gravity will rotate it downward. In this question we position the rod vertically so that the pivot is at the bottom and release it from rest. What is its angular velocity once it has rotated through an angle of 55 degrees? (In its starting position, the angle was zero degrees.) For this rod, I = ML2/3.

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In the previous question, what is the rod's angular acceleration once it has
rotated through an angle of 55 degrees? (In its starting position the angle
was zero degrees.)
O 3.51 rad/s?
O 8.03 rad/s2 ht
O 5.02 rad/s?
O 9.80 rad/s2
O 14.0 rad/s2
Transcribed Image Text:In the previous question, what is the rod's angular acceleration once it has rotated through an angle of 55 degrees? (In its starting position the angle was zero degrees.) O 3.51 rad/s? O 8.03 rad/s2 ht O 5.02 rad/s? O 9.80 rad/s2 O 14.0 rad/s2
Question 5
A thin uniform rod 2.40 m long weighing 135 N has a frictionless pivot at
one end, allowing it to rotate in a vertical plane. If it is rotated so that it is
vertical (with the pivot at the bottom) and released, the force of gravity will
rotate it downward. In this question we position the rod vertically so that
the pivot is at the bottom and release it from rest. What is its angular
velocity once it has rotated through an angle of 55 degrees? (In its starting
position, the angle was zero degrees.)
For this rod, I = ML2/3.
%3D
Transcribed Image Text:Question 5 A thin uniform rod 2.40 m long weighing 135 N has a frictionless pivot at one end, allowing it to rotate in a vertical plane. If it is rotated so that it is vertical (with the pivot at the bottom) and released, the force of gravity will rotate it downward. In this question we position the rod vertically so that the pivot is at the bottom and release it from rest. What is its angular velocity once it has rotated through an angle of 55 degrees? (In its starting position, the angle was zero degrees.) For this rod, I = ML2/3. %3D
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