How Fast Does the Center of Mass of a Falling Stick Move?

[Tanya Sharma]

Homework Statement

A stick of mass M and length L is held vertically on a frictionless surface. A gentle push at the top causes the rod to fall from this position . Find the speed of its center of mass when it makes an angle θ with vertical .

a) Use force method
b) Use energy method

Homework Equations

F=Ma
τ=Iα

The Attempt at a Solution

First I tried using the Newtons laws of translation and rotation .

1) The forces acting in the vertical direction are Mg and Normal from the floor at the lower end (N) .

$$ Mg - N = Ma_{cm} $$

2) Writing torque equation about CM ,

$$ N\frac{L}{2}sinθ = \frac{ML^2}{12}\ddotθ $$

3) y_cm = (L/2)cosθ

So , $$ v_{cm} = -\frac{L}{2}(\dotθsinθ) $$

and , $$ a_{cm} = -\frac{L}{2}[\ddotθsinθ + (\dotθ^2)cosθ] $$

Using the above three equations,I end up with

$$ a_{cm} = \frac{1}{1-3sin^2θ}(-3gsin^2θ - \frac{L}{2}\dotθ^2cosθ ) $$

But I need to find relation between v_cm and θ .

I would be grateful if somebody could help me with the problem .

 

[voko]

There was a thread about this problem a couple of days ago, where the same (difficult) approach was used.

I suggest using energy.

 

[ehild]

You can write up a_cm also in terms of theta and derivatives. Then apply the trick that d²θ/dt²= dω/dt = (dω/dθ) ω =0.5 d(ω²)/dθ. You can consider ω² a new variable, say z, and you get a first order linear equation for it (hopefully) .

ehild

 

[Tanya Sharma]

You can write up a_cm also in terms of theta and derivatives.
ehild

Is it something different from $$ a_{cm} = -\frac{L}{2}[\ddotθsinθ + (\dotθ^2)cosθ] $$ ?

Are the three equations I have formed correct and relevant ?

 

[ehild]

I meant to write a de for theta.

ehild

 

[Saitama]

Hello Tanya!

You can look at the following thread, it discusses a similar problem:

https://www.physicsforums.com/showthread.php?t=718175

 

[haruspex]

There was a thread about this problem a couple of days ago, where the same (difficult) approach was used.

I suggest using energy.

As I read the OP, the requirement is to solve it twice, once using energy and once using forces.

 

[Tanya Sharma]

I meant to write a de for theta.

ehild

Okay ...this is what i get

$$\frac{L}{6sinθ}(3sin^2θ-1)\ddotθ + \frac{L}{2}cosθ\dotθ^2 + g = 0 $$

Putting θ ^'' = ωdω/dθ and θ^' = ω

$$ \frac{L}{6sinθ}(3sin^2θ-1)ω\frac{dω}{dθ} + \frac{L}{2}cosθω^2 + g = 0 $$

Is it correct ?

 

[nil1996]

are you considering the translational velocity of the center of mass?
I am too trying to solve this problem.

 

[Tanya Sharma]

are you considering the translational velocity of the center of mass?
I am too trying to solve this problem.

v_cm is the velocity of center of mass .

 

[nil1996]

I mean that the rod as a whole is also moving in horizontal direction(as there is no friction). I don't see you have considered this velocity.

 

[voko]

Why would it move horizontally?

 

[Tanya Sharma]

I mean that the rod as a whole is also moving in horizontal direction(as there is no friction). I don't see you have considered this velocity.

Please wait for some time before you put forward any query or suggestions...

 

[ehild]

Homework Statement

A stick of mass M and length L is held vertically on a frictionless surface. A gentle push at the top causes the rod to fall from this position . Find the speed of its center of mass when it makes an angle θ with vertical .

a) Use force method
b) Use energy method

Homework Equations

F=Ma
τ=Iα

The Attempt at a Solution

First I tried using the Newtons laws of translation and rotation .

1) The forces acting in the vertical direction are Mg and Normal from the floor at the lower end (N) .

$$ Mg - N = Ma_{cm} $$

2) Writing torque equation about CM ,

$$ N\frac{L}{2}sinθ = \frac{ML^2}{12}\ddotθ $$

3) y_cm = (L/2)cosθ

I miss a picture...

If the origin is at the ground just below the CM, the y-axis points upward, and then ma_cm=d²y_cm/dt²=N-mg.
Using a_cm and v_cm is confusing, use the derivatives of y, and check the signs.

At the end, use ω² as variable, function of theta.

ehild

 

[ehild]

Okay ...this is what i get

$$\frac{L}{6sinθ}(3sin^2θ-1)\ddotθ + \frac{L}{2}cosθ\dotθ^2 + g = 0 $$

Putting θ ^'' = ωdω/dθ and θ^' = ω

$$ \frac{L}{6sinθ}(3sin^2θ-1)ω\frac{dω}{dθ} + \frac{L}{2}cosθω^2 + g = 0 $$

Is it correct ?

Check the signs. Read my previous post.

ωdω/dθ=0.5 d(ω²/dθ). You have a first order equation for ω², easy to solve.

ehild

 

[Tanya Sharma]

I miss a picture...

Here is the picture

 

[ehild]

Ok, I have the same picture.

ehild

 

[nil1996]

Why would it move horizontally?

If it is going in circular motion it will experience a radially outward force.Isn't it?So i think due to this force the center of mass will have a translational motion too as there is no friction.Is that right??

 

[Tanya Sharma]

$$ \frac{(3sin^2θ+1)}{(12sinθ)}\frac{dz}{dθ} + \frac{L}{2}(cosθ) z - g =0 $$

Does this makes sense ?

 

[ehild]

$$ \frac{(3sin^2θ+1)}{(12sinθ)}\frac{dz}{dθ} + \frac{L}{2}(cosθ) z - g =0 $$

Does this makes sense ?

You miss an L in the first term. Otherwise it is OK :thumbs:

ehild

 

[voko]

If it is going in circular motion it will experience a radially outward force.Isn't it?So i think due to this force the center of mass will have a translational motion too as there is no friction.Is that right??

If there is no friction, the net horizontal force is zero. So the horizontal momentum is conserved. We can choose an inertial reference frame where the rod has no horizontal motion.

 

[nil1996]

If there is no friction, the net horizontal force is zero. So the horizontal momentum is conserved. We can choose an inertial reference frame where the rod has no horizontal motion.

thanks i got it.:thumbs:

 

[nil1996]

$$ \frac{(3sin^2θ+1)}{(12sinθ)}\frac{dz}{dθ} + \frac{L}{2}(cosθ) z - g =0 $$

Does this makes sense ?

Sorry for interrupting you thread

 

[Tanya Sharma]

ehild...By applying method of separation of variables ; z =u(θ)v(θ)

v(θ) = C₁/(3sin²θ + 1)

u(θ) = (-12gcosθ)/(LC₁) + C₂

This doesn't give me correct answer ...Is this what you are getting or something else?

 

[ehild]

ehild...By applying method of separation of variables ; z =u(θ)v(θ)

v(θ) = C₁/(3sin²θ + 1)

u(θ) = (-12gcosθ)/(LC₁) + C₂

This doesn't give me correct answer ...Is this what you are getting or something else?

One integration constant is enough, you can take C1=1.

It looks correct, although I did not check the constants. So uv=ω², and ω=0 at θ=0. Find the constant, then calculate ω at θ=pi/2.

ehild

 

[Tanya Sharma]

One integration constant is enough, you can take C1=1.

It looks correct, although I did not check the constants. So uv=ω², and ω=0 at θ=0.

ehild

Constant C = 12g/L

ω² = 12g(1-cosθ)/[L(3sin²θ+1)] .This doesn't seem to be right.

Find the constant, then calculate ω at θ=pi/2.

Why ? What has θ=pi/2 to do ?

 

[ehild]

Sorry, I thought you need the final velocity. Why do you think it is not right?

ehild

 

[Tanya Sharma]

So, $$ ω^2 = \frac{12g(1-cosθ)}{L(3sin^2θ+1)} $$ looks alright to you ?

And then

$$ v_{cm} = -\frac{L}{2}(\dotθsinθ) $$

gives $$ ω = -[\frac{3gL(1-cosθ)(sinθ)}{(3sin^2θ+1)}]^\frac{1}{2} $$

Is this fine ?

 

[ehild]

I got the same result with the energy method.

 

[Tanya Sharma]

Why are we getting a minus sign ?

 

[ehild]

So, $$ ω^2 = \frac{12g(1-cosθ)}{L(3sin^2θ+1)} $$ looks alright to you ?

I like it

And then

$$ v_{cm} = -\frac{L}{2}(\dotθsinθ) $$

gives $$ ω = -[\frac{3gL(1-cosθ)(sinθ)}{(3sin^2θ+1)}]^\frac{1}{2} $$

Is this fine ?

why minus? And you miss a factor of 2. $$ ω =2\sqrt{ \frac{3g(1-cosθ)}{L(3sin^2θ+1)} }$$

and Vcm=-ωsinθ L/2.

 

[Tanya Sharma]

Oh ! that was a typo..I meant

$$ v_{cm} = -[\frac{3gL(1-cosθ)(sinθ)}{(3sin^2θ+1)}]^\frac{1}{2} $$

Minus makes perfect sense...

 

[ehild]

Oh ! that was a typo..I meant

$$ v_{cm} = -[\frac{3gL(1-cosθ)(sinθ)}{(3sin^2θ+1)}]^\frac{1}{2} $$

Minus makes perfect sense...

$$ v_{cm} = -[\frac{3gL(1-cosθ)(sin^2θ)}{(3sin^2θ+1)}]^\frac{1}{2} $$
sinθ has to be squared if you pull it under the square root. But otherwise it is perfect:thumbs: Have a good rest, you deserve it.

ehild

 

[Tanya Sharma]

ehild...I am falling short of words to express my gratitude...All I will say is "THANK YOU ehild" .

You rock!

 

[ehild]

ehild...I am falling short of words to express my gratitude...All I will say is "THANK YOU ehild" .

You rock!

You are welcome. That was a challenging problem. I am also exhausted

ehild