# Fixed lever arm spinning under gravity, why am I getting these results?

Suppose there is a lever arm of length $L$, a mass $m$, and it is fixed at one end. The lever is parallel to the ground. So the force acting on the center of mass of the lever would be $mg$.

Now let's look at the moment when the lever is released.

Equating the equations for torque, we have $Fr=I \alpha$, so in this case $\dfrac {mgL}{2}=\dfrac {ML^2 \alpha}{3}$ , or $\alpha=\dfrac {3g}{2L}$ . To find the tangential acceleration at the center of mass, we can use the formula $a=r \alpha$, so in this case $a= \dfrac {3g}{4}$, so $ma=\dfrac {3mg}{4}$. Why doesn't $ma$ equal $mg$?

• Why should it be? Commented Jan 13, 2014 at 9:49
• @IgnacioVergaraKausel Ahhhh thanks I think I got it, $ma$ should be the sum of ALL the forces on the lever, so $ma$ should be $mg-$ the force in the pin which is holding up the lever.
– Ovi
Commented Jan 13, 2014 at 9:52
• Possible duplicate: physics.stackexchange.com/q/74313/392 Commented Jan 13, 2014 at 17:04

In actuality there would also be a normal force at the hinge, when you apply the equation for torque from the hinge, the torque of this normal force becomes zero, but it is there nevertheless.

Your calculations are correct, if you calculate the force on rod either by integrating all forces or finding acceleration of centre of mass and multiplying by total mass in both cases the net force comes out to be ${3mg}/4$ this force comes out to be less than $mg$ because there is a normal reaction at hinge.

The force on the support against gravity goes towards rotationaly accelerating the rod. If $m a = m g$ then support force equals gravity and the rod center of mass would remain static.
PS. There was a very similar question recently here about a rod suspended by two string at the ends and finding the tension on one string the moment the other string its cut. The force was $T = \frac{1}{4} m g$ which is the same result you got. Read the discussion below for more details.