Wednesday

April 1, 2015

April 1, 2015

Posted by **roni** on Saturday, November 27, 2010 at 6:14pm.

- physics please help -
**drwls**, Saturday, November 27, 2010 at 6:23pmThe initial kinetic energy of the bar, which is

(1/2) I wo^2 = (1/2)*M (L^2/3)*(vo/L)^2

= (1/6)*M vo^2

Must equal the increase in potential energy in the upside-down position, which is

M g L

since the center of mass raises by an amount L.

I is the moment of inertia of the rod about the pivot at the top.

M cancels out and you are left with

(1/6)vo^2 = g*L

Solve for vo

**Answer this Question**

**Related Questions**

Physics - One end of a thin uniform rod, 1.00m in length, is attached to a pivot...

physics - The figure shows an overhead view of a uniform 2.00-kg plastic rod of ...

Physics - The left end of a 1-kg, 1.0 m long uniform thin rod is attached to a ...

Physics - The figure below shows a thin rod, of length L and negligible mass, ...

Physics - A 2.5 m rod of mass M= 4. kg is attached to a pivot .8m from the left ...

physics - A thin non-uniform rod of length L=2.00 m and mass M=9.00 kg is free ...

Physics - A long thin rod of mass M = 2:00 kg and length L = 75:0 cm is free to ...

Physics- Angular Momentum - A long thin rod of mass M=2.00 kg and length L=75.0 ...

Physics - A uniform rod of length L and mass m initially at rest is struck by a ...

Physics/Math - A thin rod, of length L and negligible mass, that can pivot about...