# Thin rod standing upright tips over

• cler
In summary, the conversation discusses the problem of finding the angular velocity of a uniform thin rod when it hits a fixed end after being given a small kick. The conversation includes equations and attempts at a solution using conservation of energy and the kinetic energy of a rigid body rotating around a fixed axis. It is pointed out that the rod is not a single point mass but is made up of infinite "dm" at different heights from the table. The correct method for calculating the kinetic energy is also clarified.

## 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.

## The Attempt at a Solution

So I applied conservation of energy.
Ei=Ef (1)
Ei=mgl/2
Ef=1/2mvcm2 +1/2Iω2
where vcm is the linear velocity of the CM and ω is the angular velocity of the rod when it hits the floor. vcm=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
ω2=12/7g/l

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

where am i wrong?

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

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

azizlwl said:
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.

ehild said:
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!