# Falling Rod

1. Oct 22, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data
A thin rigid rod of length $l$ is placed perfectly vertical on a smooth ground. A slight disturbance on the upper end of rod causes lower end of rod to slip along the ground and rod starts falling down. Find velocity of center of mass of rod when the rod makes $30$ with horizontal.

2.Relevant equations

3. The attempt at a solution
I first tried to find the instantaneous axis of rotation--

So distance of IAOR from Com of rod is $l cos\theta /2$

Now by conservation of energy

$\frac { mgl(1-sin\theta ) }{ 2 } =\frac { 1 }{ 2 } \left( \frac { m{ l }^{ 2 } }{ 12 } +\frac { m{ l }^{ 2 }{ cos }^{ 2 }\theta }{ 4 } \right) { \omega }^{ 2 }$

Also ${ V }_{ CM }=\omega \frac { lcos\theta }{ 2 }$

From these two equations I got ${ V }_{ CM }=\sqrt { \frac { 9gl }{ 26 } }$

I have confusion that why the lower end of the rod started moving backward.
I know that as only vertical force mg acts on the rod so the CoM of the rod falls vertically.
But why the lower end of the rod sliped backward?

Last edited: Oct 22, 2014
2. Oct 22, 2014

### ehild

If the COM falls vertically and the rod rotates how do the upper end and lower end move?

3. Oct 22, 2014

### Satvik Pandey

So does the lower end of the rod slipped backward in order to allow the CoM of the rod to fall vertically?

4. Oct 22, 2014

### ehild

Yes, and the upper end turns forward. You can describe the motion as the vertical falling of the COM and rotation about the COM.

5. Oct 22, 2014

### Satvik Pandey

I tried to draw the line of action of the velocity of the upper end of the rod.

Here the red line shows the velocity of the upper end of the rod due to rotation. The red line is perpendicular to the to the black line.

And blue line shows the velocity of upper end of the rod due to vertical fall of CoM.

And green vector is the resultant of blue and red vector.
Does green vector correctly shows the line of action of the velocity of the upper end of the rod?

6. Oct 22, 2014

### ehild

Yes, in principle. You get the velocity of the ends as the vector sum of the velocity of the COM and the velocity of the end of rod with respect to the COM.

7. Oct 22, 2014

### Satvik Pandey

I also solved this question without using IAOR.

By conservation of energy

$\frac{mgl(1-sin\theta)}{2}=\frac{mv^{2}}{2} + \frac{I \omega^{2}}{2}$

Velocity of the lower end of the rod in vertical direction is 0

So $V_{com}=\frac{l \omega cos\theta}{2}$

Using these two equations I got

$V=\sqrt { \frac { 3{ cos }^{ 2 }\theta (1-sin\theta )gl }{ 3{ cos }^{ 2 }\theta +1 } }$

Is it right?

Last edited: Oct 22, 2014
8. Oct 22, 2014

### Satvik Pandey

When we have to draw the line of action of the velocity(acceleration) of the CoM of a rigid body we first shift all the forces acting on the rigidbody to its CoM and then we find the resultant of the forces. Does the line of action of velocity(acceleration) is similar to the direction of the resultant force acting on the CoM of the body?

9. Oct 22, 2014

### ehild

Correct!

10. Oct 22, 2014

### ehild

I know what is the line of action of a force but I do not know what you mean on the line of action of velocity.

In case of the motion of a rigid body in a plane, the motion is equivalent to the motion of the CoM under the effect of all the external forces, applied at the CoM, accompanied with a rotation about the CoM.

11. Oct 23, 2014

### Satvik Pandey

I think Line of action of velocity is not the correct word.

Suppose there is rod resting in a wedge.(as shown in figure 1). When the system is released it will took the shape somewhat similar to (figure 2)(assume all surfaces to be friction less).Suppose we are asked to find the direction in which the CoM of rod will move.

I think first I have to draw the FBD of the rod.(please see figure 3)

Then I shifted all forces to the CoM of the rod.
Clearly mg is greater than N2.

Does the green arrow correctly show the direction in which CoM of rod will move?

Green vector is resultant of the three forces (mg,N1,N2).

12. Oct 23, 2014

### ehild

N1 is wrong.

13. Oct 23, 2014

### Satvik Pandey

Oh! Its direction is wrong. It should be perpendicular to the contact surface.

Does green arrow correctly shows the direction in which the CoM of the the rod will move.

Green vector is the resultant of the forces N1,N2 amd mg.

14. Oct 23, 2014

### ehild

I do not see why do you draw those arrows, but yes, the CoM moves downward and to the right. And I do not think that mg is smaller than N2 .

15. Oct 23, 2014

### Satvik Pandey

As the CoM of the rod moves downward so mg should be greater than N2.

Does the CoM of the rod moves in the resultant of N1,N2 and mg?

16. Oct 23, 2014

### ehild

Yes.

17. Oct 24, 2014

### Satvik Pandey

Thank you ehild.
You are awesome.