# Find The Percent Change In K.E. Of Rod Hit By Putty

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1. Jan 11, 2017

### XanMan

I'm trying to solve this problem, however I'm unsure if my reasoning is correct - I do agree with the answer, but something seems wrong to me in my reasoning.

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

A rod is on a frictionless surface, and is defined by the length L and mass M. It is hit on one end by a piece of putty, also of mass M, with a velocity V. The putty sticks to the rod and there is an inelastic collision between the two (of very short duration). What is the percentage of Kinetic Energy is lost during the collision?

The question has 3 more parts before the percentage bit. I shall highlight them briefly:
i) Find the velocity of the CoM before and after the collision: $\frac{V}{2}$
ii) Find the angular momentum about the CoM before the collision: $L = \frac{MVL}{4}$
iii) Find the angular velocity about the CoM after the collision: $\omega = \frac{6V}{5L}$

I worked out iii) using the conservation of angular momentum, which I am unsure if it is the correct way. The rod has no form of pivot, it will rotate about its centre of mass. However it will also have linear motion. Due to the inelastic collision, I know that linear momentum is conserved. However, is angular momentum conserved?

2. Relevant equations

I found the CoM to be located at $\frac{3L}{4}$
$$Total KE = Linear KE + Rotational KE$$
$$Linear KE = \frac{MV^2}{2}$$
$$Rotational KE = \frac{I \omega ^2}{2} = \frac{L \omega}{2}$$

3. The attempt at a solution

Before collision there is no Rotational KE. Moreover, only the putty is in motion in the system. Thus the KE Before is found using the mass of the putty and its velocity.

$$KE_{Before} = \frac{MV^2}{2}$$

The KE After is a combination of Linear and Rotational KE. Let's work them out in two separate parts.

a) Linear KE:

The velocity of the CoM after collision is $\frac{V}{2}$ and the combined mass is $2M$. Thus:

$$Linear KE = \frac{MV^2}{4}$$

b) Rotational KE:

We must first find the moment of inertia about the CoM. The moment of inertia of the system about the other end of the rod is:

$$I = I_{Rod} + I_{Putty} = \frac{ML^2}{3} + ML^2 = \frac{4ML^2}{3}$$

By Parallel Axis Theorem:

$$I_{CoM} = I - M_T X_{CoM}^2 = \frac{4ML^2}{3} - \frac{18ML^2}{16} = \frac{5ML^2}{24}$$

Finding the Rotational KE:

$$Rotational KE = \frac{I \omega ^2}{2} = \frac{3MV^2}{20}$$

Thus:

$$KE_{After} = \frac{MV^2}{4} + \frac{3MV^2}{24} = \frac{2MV^2}{5}$$

We're almost there! Huzzah!

By finding the difference between the KE Before and KE After, one sees that the percentage difference is 20%.

Last edited by a moderator: Jan 11, 2017
2. Jan 11, 2017

### Staff: Mentor

Definitely. There are no external torques, so angular momentum about the CoM will be conserved.

Your solution looks good to me.

3. Jan 11, 2017

### XanMan

Thanks Doc Al! All the best for the new year.