Thin rod standing upright tips over

[cler]

Homework Statement

A uniform thin rod of mass m and length l is standing upright on a table, then given a tiny kick so it tips over. The lower end is fixed. Find angular velocity when it hits the table.

Homework Equations

The Attempt at a Solution

So I applied conservation of energy.
Ei=Ef (1)
Ei=mgl/2
Ef=1/2mv_cm² +1/2Iω²
where v_cm is the linear velocity of the CM and ω is the angular velocity of the rod when it hits the floor. v_cm=ωl/2
For a thin rod that rotates about one end
I=ml²/3
So substituting in equation (1) i get
mgl/2=1/2(ωl/2)² + 1/6m(lω)²
which leads to
ω²=12/7g/l

But i 've checked the answer and it turns out is ω²=3g/l

where am i wrong?
thanks in advance for your help

 

[azizlwl]

It is not a single point of mass at the centre.
It is made up of infinite "dm" at different height from the table.

 

[ehild]

The kinetic energy of a rigid body rotating around a fixed axis is simply 0.5 Iω², where I is the moment of inertia with respect to the axis.

You can also calculate the KE of a rigid body moving in plane as the KE of the CM + energy of rotation around the CM. In this case, you have to use the moment of inertia with respect to the CM.

You mixed the two methods. As the bottom end of the rod is fixed you can consider it a fixed axis.

ehild

 

[cler]

It is not a single point of mass at the centre.
It is made up of infinite "dm" at different height from the table.

Ok, but basically what i do is saying that the energy at the beginning Ei is potential energy due to gravity. Then when the rod hits the table potential energy equals zero, so i only have kinetic energy which is a sum of translational and rotational energy. which is the translation of the center of mass plus the rotational energy of the body.

 

[cler]

The kinetic energy of a rigid body rotating around a fixed axis is simply 0.5 Iω², where I is the moment of inertia with respect to the axis.

You can also calculate the KE of a rigid body moving in plane as the KE of the CM + energy of rotation around the CM. In this case, you have to use the moment of inertia with respect to the CM.

You mixed the two methods. As the bottom end of the rod is fixed you can consider it a fixed axis.

ehild

Ok, now i get it. Thank you very much!