Falling Velocity Relative to Weight

[coolguy9]

Homework Statement

A wooden rod of negligible mass and length 80.0 cm is pivoted about a horizontal axis through its center. A white rat with mass 0.490 kg clings to one end of the stick, and a mouse with mass 0.240 kg clings to the other end. The system is released from rest with the rod horizontal.

If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

Homework Equations

free fall acceleration = 9.80 m/s^2.

The Attempt at a Solution

I would love to attempt at the solution, but I haven't even the slightest clue where to start, I'm quite confused :(

 

[Dick]

I would think about conservation of energy. The total kinetic energy of the unfortunate mammals is equal to the change in gravitational potential energy.

 

[coolguy9]

I'm not making too much headway on this problem..

I've got

PE = change in mgh and
KE = change in 1/2mvf^2 - 1/2mvi^2

PE = 5.7232, assuming m = 0.490 kg + .240 kg, and g = 9.8, and h = .80m

but I don't really know where to go from there. and is that PE even correct? or do I have to do it separately for each animal, and add them? ahh, so confused

 

[Dick]

Ok, so mouse goes up, rat goes down. PE=mgh. One h is positive and one is negative. Since one goes up and one goes down. Hence, don't add the masses. Calculate each one separately. Initial KE is zero.

 

[coolguy9]

Alright, I'm still a little confused but I'm getting there, I think..

PE(mouse) = (9.8m/s)(.24kg)(0.8m) = 1.8816
PE(rat) = (9.8)(.49)(-0.8) = -3.8416

change in PE = 5.7232

5.7232 = 1/2 mv^2

v = 3.38?

Which is still wrong, so.. I'm doing something wrong hehe

 

[Dick]

You have to ADD the PE's. If the mouse and the rat had the same mass there would be no change in PE.

 

[coolguy9]

Ok, so..


Total PE = -1.96

-1.96 = (1/2) (0.49 + 0.24)(v^2)

Assume the negative sign is negligible?

v = 2.3?

 

[Dick]

The minus sign isn't negligible. You just have to think about what it means relative to your choice of the sign of h. But yes, roughly 2.3m/s.

 

[coolguy9]

The website is telling me 2.3 is incorrect.. I re-calculated, and it's coming up 2.3172, rounded to the tenths is 2.3..

Is there anything else we might not be accounting for?

 

[Dick]

I don't think so. I worked it out again and agree with your number. Did you put correct units on it?