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jhson114
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https://tychosrv-s.phys.washington.edu/cgi/courses/shell/common/showme.pl?courses/phys121/spring05/homework/08/pivoting_rod_RSD/1.gif
A uniform rod is pivoted at its center and a small weight of mass M = 5.15 kg is rigidly attached to one end. The rod has length L = 4 m and mass mrod = 12.4 kg. The system is released from rest at the q = 37° angle. The angular acceleration just after it is released is 2.17rad/s^2. What is the angular velocity when the rod is vertical?
To do this, you can use the energy conservation. Change in potential energy of the rod plus the change in potential energy of the weight equals the kinetic energy. KE = (1/2)IW^2 = delta total Potentional energy. You get the change in potential energy by finding the delta PE of weight and the rod. delta PE of weight is simply M*G*(change in Height), but how do you find the change in PE of the rod?
A uniform rod is pivoted at its center and a small weight of mass M = 5.15 kg is rigidly attached to one end. The rod has length L = 4 m and mass mrod = 12.4 kg. The system is released from rest at the q = 37° angle. The angular acceleration just after it is released is 2.17rad/s^2. What is the angular velocity when the rod is vertical?
To do this, you can use the energy conservation. Change in potential energy of the rod plus the change in potential energy of the weight equals the kinetic energy. KE = (1/2)IW^2 = delta total Potentional energy. You get the change in potential energy by finding the delta PE of weight and the rod. delta PE of weight is simply M*G*(change in Height), but how do you find the change in PE of the rod?
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