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monkeysmine
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Homework Statement
A uniform rod 1.1 m long with mass 0.7 kg is pivoted at one end, as shown in Fig. 9-46, and released from a horizontal position.
Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod makes with the horizontal direction. (Use theta for θ.)
http://www.webassign.net/fgt/9-43.gif
Homework Equations
T = rFsin(theta)
The Attempt at a Solution
My attempt has consisted of the following:
1st attempt:
Using the basic formula, and then applying it to obtain ~7.6sin(theta), trying the negative version as well.
I think realized that sin might not be valid to use due to the nature of the direction of application of the force and tried 7.6cos(90-theta), which also did not work.
2nd attempt:
T = angular acceleration * moment of inertia
I solved the moment of inertia to be ~.2823 (mL^2/3).
I defined omega as d/(dt)[int(g)sin(theta)/r].
I then plugged in the values and came up with .2823d/(dt)[int(9.8)sin(theta)/1.1]
Any help would be appreciated.