# Question about motion of rod after collision

1. Dec 28, 2006

### phalanx123

Hi I am kind of stuck on this question.

Show that the moment of inertia of a uniform thin rod of mass M and length l about an axis perpendicular to the rod through its centre is
$$\frac{Ml^2}{12}$$
What is the moment of inertia of the rod about a parallel axis a distance x from the centre?

The rod defined above lies at rest on a smooth horizontal table. A point
mass m = 1/2M moving at right angles to the rod with a speed v collides with one end of the rod, and sticks to it.
(a) Describe qualitatively, including diagrams, the subsequent motion of the
rod.

(b) Derive the speed and direction of motion of the centre of mass (of the
rod and mass combined), after the collision.

(c) Derive the angular speed of the combined rod and mass after the
collision.

(d) Find the position on the rod that is stationary immediately after the
collision. [

Ok here is my attempt to the solution:
I did my derivation of the moment of inertia no problem. and using the parallel axis theorem, the new moment of inertia is I=Mx^"+Io where Io is the moment of inertia through the centre of mass of the rod.

a) the centre of mass of the rod will move forward and the rod will also rotate about the new centre of mass which is a distance x away from the original i.e. at l/2

b)Using conservation of momentum
mv=(m+M)v1 which gives 1/2Mv=(3/2)Mv1
so v1=1/3v

Part C is where I am begining to get stuck.

c) using conservation of energy and angular momentum

$$\frac{1}{2}(\frac{1}{2}M)v^2=\frac{3}{2}M(\frac{1}{3}v)^2+\frac{1}{2}I\omega^2$$

this gives $$\frac{1}{6}Mv^2=\frac{1}{2}I\omega^2$$

from angular momentum
1/2Mv(l/2-x)=I*omega where x is the distance between th new and old centre of mass

from this Mv =(2*I*omega)/(l/2-x) and substitute this into the previous equation gives 1/6*(2*I*omega)v/(l/2-x)=1/2I*omega^2

this gives omega=2/3 v/(l/2-x)

d) substitute omega=2/3 v/(l/2-x)back into 1/2Mv(l/2-x)=I*omega gives
I=3/4M(l/2-x)^". this is the moment of inertia of rod and poin mass combined about the axis which passes through the new centre of mass

the moment of inertia Io of combined masses through the initial centre of mass is Io=Ml^2/12+1/2m(l/2)^2=Ml^2/12+Ml^2/16=7Ml^2/48

using parralel axis theorem Io=Mx^2+I
solving the quadratic equation gives x=(3+/-sqrt(3))l/18

I really don't know if I am on the right track or not , could somebody help me please. thanks

Last edited: Dec 28, 2006
2. Dec 28, 2006

### Staff: Mentor

This collision is perfectly inelastic--energy is not conserved. But angular momentum is conserved. Hint: Where is the center of mass of the combined object? What is its moment of inertia about that point? What is the angular momentum about that point?

3. Dec 28, 2006

### phalanx123

This is actually where I am stuck on. I don't really know how to calculate the centre of mass/moment of inertial of a ununiformed object. I am thinking of using integration. But that is for uniformed object isn't it? As for the moment of inertia. I think I can use the parallel axis theorem by first calculating the moment of inertia of the combined mass about the axis at l/2 (Ml^2/12+1/2m(l/2)^2=Ml^2/12+Ml^2/16=7Ml^2/48 is this right?) than by setting the distance between this point and the centre of mass to be X the moment of inertia at the CM is then I=2/3MX^2+7Ml^2/48. But I will have to introduce this variable X and further complictae the question.

4. Dec 28, 2006

### Hootenanny

Staff Emeritus
Start by finding the centre of mass of the composite body. Then, find the moment of inertia of the body about this axis (as Doc Al suggests).

Last edited: Dec 28, 2006
5. Dec 28, 2006

### Staff: Mentor

There's no need for integration. If you have two point masses, how would you find their center of mass? (For the purpose of calculating the center of mass of the system, you can replace the stick by an equivalent mass located at the stick's center of mass.)

Not quite sure I follow what you are doing here. As I said earlier: Find the center of mass of the system, then find the moment of inertia of the system about that point. (You'll need the parallel axis theorem to find the stick's moment of inertia about the system center of mass.)

6. Dec 28, 2006

### phalanx123

Ok I got it now. Thanks a lot ^_^