# Rod colliding with the particle.

1. Mar 28, 2015

### Satvik Pandey

1. The problem statement, all variables and given/known data
A rod of length 2 m($l$. and mass 16 kg ($M$) is projected vertically upwards from the ground, such that the top of the rod reaches a height $h$. At the moment that it reaches its apex, a ball of mass 3 kg($m$ hits the top of the rod with a velocity of 10 m/s($v$) and binds to the very end of the rod. The rod, now starts to spin. It perfectly completes one full rotation, and then hits the ground. Find $h$.

2. Relevant equations

3. The attempt at a solution

As no external torque was acting on the system just before the collision so I think I can conserve angular momentum of the system about point $O$.

Initial momentum will be $0.5mvl=30$

After collision the ball sticks to the rod so position of CoM will change. Let the CoM be located at $x$ distance from point O.

So $x=\frac{ml}{2(M+m)}=\frac{3}{19}$

Now final can angular momentum can be find by adding the angular momentum of the CoM about the point O and angular momentum of the system about it's CoM.

So Final angular momentum = $(M+m)v_{0}x+I_{CoM} \omega$

As no external force acts on the system in horizontal direction so we can conserve linear momentum.

$mv=(M+m)v_{0}$

Now $I_{CoM}$= I{rod about CoM}+I {m about CoM}

=$\frac { M{ l }^{ 2 } }{ 12 } +M{ x }^{ 2 }+m(1-x)^{ 2 }$

On putting values of $x$ I got it's value =$\frac { 16\times 28 }{ 3\times 19 }$

Putting this in the equation framed by conserving angular momentum I got

$\omega =\frac { 15(3) }{ 14 }$

Time taken to complete one rotation is $\frac{2 \pi}{\omega}$

So $t=\frac{88}{45}$

Now the rod will fall by distance $h-2$ before hitting the ground. So time required will be $\sqrt { \frac { h-2 }{ 4.9 } }$

Equating these I got $h =20.738$. But this is not correct? I am unable to find my mistakes. Please help me!

2. Mar 28, 2015

### Staff: Mentor

I think it is easier to consider the rotation in the center of mass system.
You use angular momentum around O, but then moment of inertia for the CoM? Those don't fit together.

3. Mar 28, 2015

### Satvik Pandey

I have found angular momentum about O. Let O be a point in a plane which coincide with the center of the rod initially. After collision the rod begins to rotate and translate together. Now angular momentum of a body about the point is found by treating the body as a point mass located at the CoM and finding the angular momentum of the CoM relative to that point and then adding the angular momentum of the body about the CoM. And angular momentum of a body about the CoM is the $I_{com} \omega_{com}$ So I have to consider moment of inertia about CoM.

4. Mar 28, 2015

### Staff: Mentor

Angular momentum cannot be relative to two points at the same time - CoM or O, but not both.

5. Mar 28, 2015

### Delta²

You should apply conservation of angular momentum before and after the collision around the same stable point, let that point be the CoM of the system. Dont use different points for before and after the collision (O before the collision, CoM after the collision) as you do now.

6. Mar 28, 2015

### Satvik Pandey

I have not found angular momentum about two points. Before collision I found angular momentum about point O. Then after collision I have also found angular momentum about point O. After collision the rod will translate and rotate simultaneously. Angular momentum about a point is found by finding the angular momentum of the CoM about that point and then adding angular momentum of the body about the CoM.
The term $(m+M)v_{0}x$ is the angular momentum of the CoM of the rod about point O and $I_{com} \omega$ is the angular momentum of the system about the CoM. Adding these gives the angular momentum of the system about point O.

7. Mar 28, 2015

### Himanshu_123

Is the correct ans. approx. 17m?

8. Mar 28, 2015

### Satvik Pandey

I don't know bro. I found this question online. So I don't have it's solution.

9. Mar 28, 2015

### Satvik Pandey

In this question please look at the second method of the solution (eq 7.53) . In this method the angular momentum is conserved about the point which initially coincides with the center of the stick. The term in LHS shows the initial angular momentum of the system about that point. The first term in the RHS shows the angular momentum (spin angular momentum) about the CoM and the last term in the RHS shows the angular momentum of the CoM of the system about that point which initially coincides with the center of the stick. Just like this I have done same in this question also. What am I doing wrong?

10. Mar 28, 2015

### TSny

I get Satvik's answer. I tried both the methods shown in the figure of post #9 and got the same answer either way.

11. Mar 28, 2015

### Satvik Pandey

Yay! my answer is right. I think that website has the wrong answer of this question. Thank you TSny. Thank you every one for helping me in this question.

12. Mar 28, 2015

### TSny

We can't assume that our answer is correct. Maybe we're both making the same mistake!

What answer did the website give?

13. Mar 28, 2015

### Satvik Pandey

It asked us to find ceiling function of $h/10$. And if $h=20.78$ then the answer should be $3$. But the answer on that website was $6$.

14. Mar 28, 2015

### TSny

Ah. Now I understand why you said that you weren't getting the website's answer even though the website didn't give an answer for h.

15. Mar 28, 2015

### Staff: Mentor

Okay, I calculated it, I get the same number for angular velocity and height (20.743m).

6 is certainly wrong.

16. Mar 28, 2015

### ehild

My h is also 20.7 m.

17. Mar 28, 2015

### TSny

OK, Satvik. I think you can now take it to the bank.

18. Mar 29, 2015

### Satvik Pandey

Thank you every one for helping me in this question.