(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform horizontal rod of mass 2.3 kg and

length 0.32 m 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 4.6 N force at an angle of 76 to the hor-

izontal acts on the rod as shown, what is the

magnitude of the resulting angular accelera-

tion about the pivot point? The acceleration

of gravity is 9.8 m/s2 .

Answer in units of rad/s2

2. Relevant equations

torque = r * F

I = (ml^^{2})/12

torque = I[itex]\alpha[/itex]

3. The attempt at a solution

I = ml^2/12

=((2.3)(0.32)^2)/12

=0.019626667

torque = r * F

= 0.32 * 4.6sin 76

=1.428275309

torque = I[itex]\alpha[/itex]

1.428275309 = 0.01962667[itex]\alpha[/itex]

[itex]\alpha[/itex]= 72.77216711

a=r[itex]\alpha[/itex]

=72.77216711 * 0.32

=23.28709348

BUT it's wrong...

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# Torque 3

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