What is the linear velocity of a point on a swinging rod before it hits a wall?

[T-Rent]

Homework Statement

If you suddenly cut the string, the rod will swing down. Find the linear velocity (V) of
the point P just before the rod hits the wall (the momentum-of-inertial I=(1/3)ML^2 if the rod rotates about the end point Q. L is the rod length and M is the rod mass).

The rod has a mass of 1kg.

A link to the picture
http://pictureserver.bravehost.com/problem_5.jpeg

Homework Equations

I=(1/3)ML^2

The Attempt at a Solution

I've found the tension on the string and the force at point Q.(.433kg and .375kg respectively).

For the linear velocity can I just find the center of mass (0.5m), treat it like a particle attached to a massless string and use v=sqrt(2*g*h)?
Or is my logic totally flawed? Any help would be great!

 

[tiny-tim]

Welcome to PF!

Hi T-Rent! Welcome to PF!

(try using the X² icon just above the Reply box )

For the linear velocity can I just find the center of mass (0.5m), treat it like a particle attached to a massless string and use v=sqrt(2*g*h)?
Or is my logic totally flawed?

Yes.

You need to use the rotational formula for kinetic energy

 

[T-Rent]

Thanks for the quick reply. I'm having some trouble understanding this one. So my KE=1/2*I*W². What would my W (angular velocity) be?

 

[tiny-tim]

Hi T-Rent!

(have an omega: ω )

That's for you to find out, from conservation of energy.

 

[T-Rent]

So, let's give this a shot.

PE=mg(L/2)=5J
Rotational KE=1/2*I*ώ²
I=1/3*M*L^2=1/3kg*m²
So solving for ώ gets me 5.46 rad/s which is the same in m/s because my L and radius is 1m.

Does this seem right?

 

[tiny-tim]

Looks good!