A uniform rod of length 1.29 m is attached to a frictionless pivot at one end. It is released from rest from an angle q = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass.(adsbygoogle = window.adsbygoogle || []).push({});

[tex] I= (1/2)ML^2 [/tex]

I= (1/2)(1.29^2), since mass isn't given I just used 1 kg.

I=.832 kg*m^2

[tex] \tau= rFsin \theta [/tex] =.645(9.8)(sin 21)= 2.27 Nm

[tex] \tau = I \alpha [/tex]

2.27= .832 [tex] \alpha [/tex]

[tex] \alpha [/tex] = 2.73 rad/s^2

[tex] a= \alpha * r [/tex]

a= 2.73 * .645

a= 1.76 m/s^2

This isn't right.. can someone help me?

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# Homework Help: Pivoting rod

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