a rod of length 1m and mass 0.5 kg hinged at one end, is initially hanging vertical. the other end is now raised slowly until it makes an angle 60 degree with the vertical. the required work is
Answers
Answered by
22
Since, this is rotational setting where the factor gravity works in the downward direction and support resists upward. For this reason, we have a torque of this scale:
Torque=(mh/2)g cos(60)
The work done in this case is calculated by integrating tangential force along the arc from 60 degrees to vertical.
Torque=(mh/2)g cos(60)
The work done in this case is calculated by integrating tangential force along the arc from 60 degrees to vertical.
Answered by
0
Answer:
5/4joule is the correct answer
Explanation:
Similar questions