Falling stick

1. May 1, 2014

toothpaste666

1. The problem statement, all variables and given/known data

A thin stick of length L is balanced vertically on frictionless ice. It then tips over and falls. How fast is the center of mass of the stick moving just before it hits the ice?

3. The attempt at a solution

using energy:

$I = \frac{1}{3}mL^2$

$.5Iω^2 = mgh$

$.5Iω^2 = mg(.5L)$

$Iω^2 = mgL$

$(\frac{1}{3}mL^2) ω^2 = mgL$

$(\frac{L}{3}) ω^2 = g$

$ω^2 = (\frac{3g}{L})$

$ω = (\frac{3g}{L})^{\frac{1}{2}}$

2. May 1, 2014

jackarms

That all looks good. I would convert from angular speed to linear speed though.

3. May 1, 2014

toothpaste666

so

v=Rω
$v = L (\frac{3g}{L})^\frac{1}{2}$

$v = (3Lg)^\frac{1}{2}$

4. May 1, 2014

goraemon

Be careful when converting - the question asks how fast the center of mass (not the edge) of the stick is moving.

5. May 1, 2014

toothpaste666

oh right ..

$v = \frac{L}{2} (\frac{3g}{L})^\frac{1}{2}$

$v = \frac{1}{2}(3Lg)^\frac{1}{2}$

6. May 1, 2014

Tanya Sharma

This doesn't look alright .The lower tip of the rod is not fixed which means the rod is not in pure rotation about the lower end point .The motion of the stick can be thought of as translation of CM+rotation about the CM .

Your approach would have been correct if the lower end point had been pivoted.

Surprisingly the end result in post#5 looks correct .

7. May 1, 2014

collinsmark

There's something else. The stick is on frictionless ice. That makes a difference compared to what it would be if the stick were on solid ground.

It means the stick does not pivot on its bottom, on the ground. Instead, the bottom end of the stick slides along the ground. Rotationally speaking, the stick pivots at its center! You'll need a different formula for the stick's moment of inertia.

8. May 1, 2014

toothpaste666

so i should use 1/12 instead of 1/3?

9. May 2, 2014

collinsmark

I think so. Do you concur? (Do you understand why I would use 1/12 instead of 1/3?)

10. May 2, 2014

toothpaste666

because its rotating about the CM and not the tip? which is also why I have to make R = L/2 instead of L?

11. May 2, 2014

toothpaste666

$I = \frac{1}{12}mL^2$

$.5Iω^2 = mgh$

$.5Iω^2 = mg(.5L)$

$Iω^2 = mgL$

$(\frac{1}{12}mL^2) ω^2 = mgL$

$(\frac{L}{12}) ω^2 = g$

$ω^2 = (\frac{12g}{L})$

$ω = (\frac{12g}{L})^{\frac{1}{2}}$

then when converting:

v= Rω

$v = \frac{L}{2} (\frac{12g}{L})^{\frac{1}{2}}$

$v = \frac{1}{2} (12gL)^{\frac{1}{2}}$

12. May 2, 2014

TSny

As Tanya said, "The motion of the stick can be thought of as translation of CM+rotation about the CM."

You've taken care of the KE due to rotation about the CM, but you have not yet included the KE due to translation of the CM.

13. May 2, 2014

toothpaste666

oh I see so not all of its energy is put into rotating because some of it needs to be used for it to slide?
so it would be
$mgh = .5Iω^2 + .5mv^2$

so i should make it so that both velocities are in terms of v and not ω?

$mgh = .5I(\frac{v}{r})^2 + .5mv^2$

$mgh = .5I(\frac{v}{\frac{L}{2}})^2 + .5mv^2$

$mgh = .5I(\frac{2v}{L})^2 + .5mv^2$

14. May 2, 2014

toothpaste666

$mgh = .5(\frac{1}{12}mL^2)(\frac{4v^2}{L^2})+ .5mv^2$

$mgh = .5(\frac{1}{12}m)(4v^2)+ .5mv^2$

$mg(.5L) = .5(\frac{1}{3}m)(v^2)+ .5mv^2$

$gL = (\frac{1}{3})(v^2)+ v^2$

$gL = (\frac{4}{3})(v^2)$

$v = (\frac{3}{4}gL)^\frac{1}{2}$

Last edited: May 2, 2014
15. May 2, 2014

collinsmark

There ya go. I think that's right.

16. May 2, 2014

TSny

That looks good.

It's important to understand why $v_{cm} = \omega \frac{L}{2}$ at the instant the stick strikes the ice. This relation is not true at other times.

17. May 2, 2014

Tanya Sharma

Hello TSny

Because as the stick is about to strike the ice ,it becomes horizontal and the speed of the lower tip reduces to zero .Hence the lower tip may be treated as an instantaneous axis of rotation .Or in other words the stick is in pure rotation about the lower tip just as it is about to strike the ice .Hence the relation vcm = ωL/2 is applicable .At other times the lower tip is moving so the relation vcm = ωL/2 doesn't hold .

Is it correct ?

Last edited: May 3, 2014
18. May 3, 2014

TSny

Yes, that's correct.

A related problem came up recently with a cube instead of a stick. https://www.physicsforums.com/showthread.php?t=750988

Here the instantaneous axis of rotation is not on the frictionless surface at the final position.

19. May 3, 2014

SammyS

Staff Emeritus
I'm just speculating here.

It seems these solutions are assuming that the end of the stick which is initially in contact with the ice -- they're assuming that end remains in contact with the ice.

I'm thinking that might not be the case.

None of the solutions includes the normal force, torque, or angular acceleration. (Unless, of course, that I missed something, which I'm quite good at doing.)

20. May 3, 2014

Thanks...