Speed of Stick's CM Just Before Impacting Ice

[toothpaste666]

Homework Statement

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?

The Attempt at a Solution

using energy:

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

$.5Iω^2 = mgh$

$.5Iω^2 = mg(.5L)$

$Iω^2 = mgL$

$(\frac{1}{3}mL^2) ω^2 = mgL$

$(\frac{L}{3}) ω^2 = g$

$ω^2 = (\frac{3g}{L})$

$ω = (\frac{3g}{L})^{\frac{1}{2}}$

 

[jackarms]

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

 

[toothpaste666]

so

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

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

 

[goraemon]

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

 

[toothpaste666]

oh right ..

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

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

 

[Tanya Sharma]

using energy:

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

$.5Iω^2 = mgh$

$.5Iω^2 = mg(.5L)$

$Iω^2 = mgL$

$(\frac{1}{3}mL^2) ω^2 = mgL$

$(\frac{L}{3}) ω^2 = g$

$ω^2 = (\frac{3g}{L})$

$ω = (\frac{3g}{L})^{\frac{1}{2}}$

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 .

 

[collinsmark]

Homework Statement

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?

The Attempt at a Solution

using energy:

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

$.5Iω^2 = mgh$

$.5Iω^2 = mg(.5L)$

$Iω^2 = mgL$

$(\frac{1}{3}mL^2) ω^2 = mgL$

$(\frac{L}{3}) ω^2 = g$

$ω^2 = (\frac{3g}{L})$

$ω = (\frac{3g}{L})^{\frac{1}{2}}$

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.

 

[toothpaste666]

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

 

[collinsmark]

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

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

 

[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?

 

[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

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


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

 

[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.

 

[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?

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

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

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

 

[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$

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

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

 

[collinsmark]

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


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


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


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

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

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

There you go. I think that's right.

 

[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.

 

[Tanya Sharma]

Hello TSny

Please check my understanding.

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.

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 v_cm = ωL/2 is applicable .At other times the lower tip is moving so the relation v_cm = ωL/2 doesn't hold .

Is it correct ?

 

[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 v_cm = ωL/2 is applicable .

Is it correct ?

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.

 

[SammyS]

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.)

 

[Tanya Sharma]

Yes, that's correct.

Thanks...

 

[haruspex]

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.)

Yes, I agree that's a valid concern. I investigated this some years ago and managed to show that it does stay in contact with the ice, but I couldn't see a trivial way to show it.
(I believe it will become airborne if there's sufficient friction.)

 

[SammyS]

Yes, I agree that's a valid concern. I investigated this some years ago and managed to show that it does stay in contact with the ice, but I couldn't see a trivial way to show it.
(I believe it will become airborne if there's sufficient friction.)

Thanks haruspex. Nice to confirm that I wasn't completely off base!

It seems to me that if you do this with normal force included, you can see if the normal force ever goes negative. (I don't know if this is trivial. ) :-)


I recall that if you have a ladder slide down a wall with both vertical and horizontal surfaces frictionless, the ladder will lose contact will one or both surfaces.

 

[toothpaste666]

thank you all!