Thin rod standing upright tips over

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A uniform thin rod of mass m and length l, initially standing upright, tips over when given a slight kick, and the problem involves finding its angular velocity upon hitting the table. The conservation of energy approach is applied, equating initial potential energy to the sum of translational and rotational kinetic energy at the moment of impact. The moment of inertia for the rod is correctly identified as I=ml²/3, but confusion arises from mixing methods of calculating kinetic energy. The correct angular velocity is derived as ω²=3g/l, clarifying the need to consider the fixed axis of rotation at the rod's base. The discussion emphasizes the importance of correctly applying energy conservation principles in rotational dynamics.
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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=1/2mvcm2 +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
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
 
Last edited:
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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!
 
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