I\'ve tried multiple ways and cannot seem to get the answer :< Question: [The ro

Question

I've tried multiple ways and cannot seem to get the answer :<

Question: [The rotational inertia of a uniform stick of mass m and length l about one end is I=(1/3)ml^2] A uniform horizontal rod of mass 1.4 kg and length 0.81m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I=(ml^2)/(12). If a 9N force at angle 73 degrees to the horizontal acts on the rod as shown, what is the magnitude of the resulting angular acceleration about the pivot point? The acceleration of gravity is 9.8m/s2. Answer in units of rad/s^2.

thank you so much!

Explanation / Answer

perpendicular forces on rod are Fsin73 and mg torque=0.81*Fsin73 -0.405*1.4*9.81=1.409 Nm torque=I*angular acc 1.409=(1/3)ml^2 *angular acc angular acc=4.601 rad/sec^2

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