A uniform horizontal rod of mass 1.8 kg and length 2.2 m is free to pivot about

Question

A uniform horizontal rod of mass 1.8 kg
and length 2.2 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 =m?^2/12.

The acceleration of gravity is 9.8 m/s2 .

If a 6.3 N force at an angle of 69? to the
horizontal acts on the rod as shown, what is
the magnitude of the resulting angular accel-
eration about the pivot point?
Answer in units of rad/s2.

I've tried this problem numerous times and I think I'm either not taking the net torque correctly or I'm using the wrong equation for I.

Explanation / Answer

net torque= 2.2Fsin69 - mg(1.1)=12.94 - 19.404 = -6.46458

I = m^2/12 + m(1.1^2) = 2.448

=-6.46458/2.448 = -2.64075 rad/s

the negative means clockwise to me but it might be positive depending on the frame of reference you usually used

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