# Torque 3

1. Oct 20, 2011

### DrunkApple

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 .

2. Relevant equations
torque = r * F
I = (ml^2)/12
torque = I$\alpha$

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$\alpha$
1.428275309 = 0.01962667$\alpha$
$\alpha$= 72.77216711
a=r$\alpha$
=72.77216711 * 0.32
=23.28709348
BUT it's wrong...

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2. Oct 20, 2011

### SteamKing

Staff Emeritus
Hint: is the pivot point of the rod located at the center of mass?

3. Oct 20, 2011

### DrunkApple

oh do i use parallel axis theorem here?