# Stick falling and slipping

1. Oct 23, 2013

### Tanya Sharma

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

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

2. Relevant equations

F=Ma
τ=Iα

3. 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) ycm = (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 vcm and θ .

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

Last edited: Oct 24, 2013
2. Oct 23, 2013

### voko

I suggest using energy.

3. Oct 24, 2013

### ehild

You can write up acm 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

Last edited: Oct 24, 2013
4. Oct 24, 2013

### Tanya Sharma

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

Are the three equations I have formed correct and relevant ?

Last edited: Oct 24, 2013
5. Oct 24, 2013

### ehild

I meant to write a de for theta.

ehild

6. Oct 24, 2013

### Saitama

7. Oct 24, 2013

### haruspex

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

8. Oct 24, 2013

### Tanya Sharma

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 ?

Last edited: Oct 24, 2013
9. Oct 24, 2013

### nil1996

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

10. Oct 24, 2013

### Tanya Sharma

vcm is the velocity of center of mass .

11. Oct 24, 2013

### 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.

12. Oct 24, 2013

### voko

Why would it move horizontally?

13. Oct 24, 2013

### Tanya Sharma

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

14. Oct 24, 2013

### ehild

I miss a picture...

If the origin is at the ground just below the CM, the y axis points upward, and then macm=d2ycm/dt2=N-mg.
Using acm and vcm is confusing, use the derivatives of y, and check the signs.

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

ehild

15. Oct 24, 2013

### ehild

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

Last edited: Oct 24, 2013
16. Oct 24, 2013

### Tanya Sharma

Here is the picture

#### Attached Files:

• ###### stick.PNG
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17. Oct 24, 2013

### ehild

Ok, I have the same picture.

ehild

18. Oct 24, 2013

### nil1996

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??

19. Oct 24, 2013

### Tanya Sharma

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

Does this makes sense ?

20. Oct 24, 2013

### ehild

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

ehild